C H a P T E R 1 0 Conics, Parametric Equations, and Polar Coordinates

Home , Semicubical parabola

CHAPTER 10 Conics, Parametric Equations, and Polar Coordinates

Section 10.1 Conics and Calculus ...... 360

Section 10.2 Plane Curves and Parametric Equations ...... 383

Section 10.3 Parametric Equations and Calculus ...... 392

Section 10.4 Polar Coordinates and Polar Graphs ...... 407

Section 10.5 Area and Arc Length in Polar Coordinates ...... 423

Section 10.6 Polar Equations of Conics and Kepler’s Laws ...... 436

Review Exercises ...... 444

Problem Solving ...... 461 CHAPTER 10 Conics, Parametric Equations, and Polar Coordinates

Section 10.1 Conics and Calculus

1. y2 4x 2. x2 8y 3. x 32 2y 2 Vertex: 0, 0 Vertex: 0, 0 Vertex: 3, 2 1 p 1 > 0 p 2 > 0 p 2 < 0 Opens to the right Opens upward Opens downward Matches graph (h). Matches graph (a). Matches graph (e).

x 22 y 12 x2 y2 x2 y2 4. 1 5. 1 6. 1 16 4 9 4 9 9 Center: 2, 1 Center: 0, 0 Circle radius 3. Ellipse Ellipse Matches (g) Matches (b) Matches (f)

y2 x2 x 22 y2 7. 1 8. 1 16 1 9 4 Hyperbola Hyperbola Center: 0, 0 Center: 2, 0 Vertical transverse axis Horizontal transverse axis Matches (c) Matches (d)

2 3 2 9. y 6x 4 2 x 10. x 8y 0 Vertex: 0, 0 y x2 42y y 3 8 Focus: , 0 (0, 0) 2 Vertex: 0, 0 x 3 −8 −4 4 8 4 Directrix: x 2 Focus: 0, 2 − (0, 0) 4 x 12 8 4 Directrix: y 2 −8 4 −12 8

11. x 3 y 22 0 12. x 12 8y 2 0

2 1 2 y 2 4 4 x 3 x 1 4 2 y 2 Vertex: 3, 2 y Vertex: 1, 2 y

Focus: 3.25, 2 4 Focus: 1, 4 x −8 −4 4 8 Directrix: x 2.75 (−3, 2) 2 Directrix: y 0 −4 x (1,− 2) −8 −6 −4 −2 −8 −2

−12 −4

360 Section 10.1 Conics and Calculus 361

13. y2 4y 4x 0 14. y2 6y 8x 25 0 y2 4y 4 4x 4 y2 6y 9 8x 25 9 y 22 41x 1 y 32 42x 2

Vertex: 1, 2 y Vertex: 2, 3 y Focus: 0, 2 6 Focus: 4, 3 8 Directrix: x 2 4 Directrix: x 0 4 (−1, 2) x −20 −16 −12 −8 −4 (2,3)−−

x −8 2 2 4 6 −12 2

15. x2 4x 4y 4 0 16. y2 4y 8x 12 0 x2 4x 4 4y 4 4 y2 4y 4 8x 12 4 x 22 41y 2 y 22 42x 2

Vertex: 2, 2 y Vertex: 2, 2 y

4 Focus: 2, 1 4 Focus: 0, 2 (−2, 2) 2 Directrix: y 3 Directrix: x 4 x −6 −4 −2 4 6 x (2,− 2) −6 −4 −2 2 −4 −2 −6 − 4 −8

2 1 2 1 2 17. y x y 0 18. y 6 x 8x 6 6 x 8x 16 10 2 1 1 2 y y 4 x 4 6y x 4 10 2 1 1 1 2 y 2 4 4 x 4 6y 10 x 4 1 1 5 2 2 Vertex: 4, 2 x 4 6 y 3 1 Focus: 0, 2 2 3 5 − x 4 4 2 y 3 1 5 2 Directrix: x 2 5 4 Vertex: 4, 3 1 ± 1 y1 2 4 x Focus: 4, 1 −3 6 1 1 19 −2 10 y2 2 4 x Directrix: y 6

−4

19. y2 4x 4 0 20. x2 2x 8y 9 0 y2 4x 4 x2 2x 1 8y 9 1 41x 1 x 12 42y 1

Vertex: 1, 0 4 Vertex: 1, 1 2 Focus: 0, 0 Focus: 1, 3 −8 10 Directrix: x 2 −6 6 Directrix: y 1 y1 2 x 1 −4 −10 y2 2 x 1 362 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

21. y 22 42x 3 22. x 12 42y 2 y2 4y 8x 20 0 x2 2x 8y 15 0

23. x h2 4py k 24. Vertex: 0, 2 x2 46y 4 y 22 42x 0 x2 24y 96 0 y2 8x 4y 4 0

25. y 4 x2 26. y 4 x 22 4x x2 x2 y 4 0 x2 4x y 0

27. Since the axis of the parabola is vertical, the form of the 28. From Example 2:4p 8 or p 2 equation is y ax2 bx c. Now, substituting the val- Vertex: 4, 0 ues of the given coordinates into this equation, we obtain x 42 8y 0 3 c, 4 9a 3b c, 11 16a 4b c. 2 5 14 x 8x 8y 16 0 Solving this system, we have a 3, b 3 , c 3. Therefore,

5 2 14 2 y 3x 3 x 3 or 5x 14x 3y 9 0.

29. x2 4y2 4 y 30. 5x2 7y2 70 y 6 x2 y2 2 x2 y2 1 1 4 4 1 14 10 2 (0, 0) 2 2 2 x 2 2 2 x a 4, b 1, c 3 −11 a 14, b 10, c 4 −6 −2 2 4 6

Center: 0, 0 Center: 0, 0 −4 − Foci: ±3, 0 2 Foci: ±2, 0 −6 Vertices: ±2, 0 Vertices: ±14, 0 3 2 14 e e 2 14 7

x 12 y 52 y x 22 y 42 y 31. 1 32. 1 9 25 12 1 14 x (1, 5) −5 −4 −3 −2 −1 − 2 2 2 1 a 25, b 9, c 16 2 2 1 2 3 a 1, b , c −2 4 4 4 (2,4)−− Center: 1, 5 −3 x Center: 2, 4 −4 Foci: 1, 9 , 1, 1 8 4 4 8 3 −5 Vertices: 1, 10 , 1, 0 Foci: 2 ± , 4 2 4 e Vertices: 1, 4, 3, 4 5 3 e 2 Section 10.1 Conics and Calculus 363

33. 9x2 4y2 36x 24y 36 0 9x2 4x 4 4y2 6y 9 36 36 36 36 x 22 y 32 1 4 9 a2 9, b2 4, c2 5 y Center: 2, 3 Foci: 2, 3 ± 5 6 Vertices: 2, 6 , 2, 0 (−2, 3)

5 2 e 3 x 6 4 2 2

34. 16x2 25y2 64x 150y 279 0 35. 12x2 20y2 12x 40y 37 0 16x2 4x 4 25y2 6y 0 279 64 225 1 12x2 x 20y2 2y 1 37 3 20 4 10 60 x 22 y 32 1 58 25 x 122 y 12 1 5 3 5 2 9 a2, , b2 , c2 a2 b2 8 5 40 a2 5, b2 3, c2 2 Center: 2, 3 1 Center: , 1 310 2 Foci: 2 ± , 3 20 1 Foci: ± 2, 1 10 2 Vertices: 2 ± , 3 4 1 Vertices: ± 5, 1 2 c 3 e a 5 Solve for y:

y 20y2 2y 1 12x2 12x 37 20 1 57 12x 12x2 2 x y 1 −1 1 234 20 −1 57 12x 12x2 −2 (2, −3) y 1 ± 20 −3

−4 (Graph each of these separately.)

−33

−3 364 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

36. 36x2 9y2 48x 36y 43 0 37. x2 2y2 3x 4y 0.25 0 4 4 9 1 9 36x2 x 9y2 4y 4 43 16 36 x2 3x 2y2 2y 1 2 4 3 9 4 4 4 9 x 322 y 12 1 4 2 x 232 y 22 1 14 1 a2 4, b2 2, c2 2 1 3 3 a2 1, b2 , c2 Center: , 1 4 4 2

2 3 Center: , 2 Foci: ± 2, 1 3 2 2 3 1 7 Foci: , 2 ± Vertices: , 1, , 1 3 2 2 2 2 2 1 Vertices: , 3, , 1 Solve for y: 2y2 2y 1 x2 3x 2 3 3 4 Solve for y: 1 7 y 12 3x x2 2 4 9y2 4y 4 36x2 48x 43 36 7 12x 4x2 36x2 48x 7 y 1 ± y 22 8 9 (Graph each of these separately.) 1 y 2 ± 36x2 48x 7 3 1 (Graph each of these separately.) −2 4

−3

−4 2

−1

38. 2 x2 y2 4.8x 6.4y 3.12 0 50x2 25y2 120x 160y 78 0 12 36 32 256 50x2 x 25y2 y 78 72 256 250 5 25 5 25 x 652 y 1652 1 5 10 a2 10, b2 5, c2 5 6 16 Center: , 5 5 6 16 Foci: , ± 5 7 5 5 6 16 Vertices: , ± 10 5 5 Solve for y: y2 6.4y 10.24 2x2 4.8x 3.12 10.24 −7 5 −1 y 3.22 7.12 4x 2x2 y 3.2 ± 7.12 4x 2x2 (Graph each of these separately.) Section 10.1 Conics and Calculus 365

39. Center: 0, 0 40. Vertices: 0, 2, 4, 2 Focus: 2, 0 1 Eccentricity: 2 Vertex: 3, 0 Horizontal major axis Horizontal major axis Center: 2, 2 a 3, c 2 ⇒ b 5 a 2, c 1 ⇒ b 3 x2 y2 1 x 22 y 22 9 5 1 4 3

41. Vertices: 3, 1, 3, 9 42 Foci: 0, ±5 Minor axis length: 6 Major axis length: 14 Vertical major axis Vertical major axis Center: 3, 5 Center: 0, 0 a 4, b 3 c 5, a 7 ⇒ b 24 x 32 y 52 x2 y2 1 1 9 16 24 49

43. Center: 0, 0 44. Center: 1, 2 Horizontal major axis Vertical major axis Points on ellipse: 3, 1, 4, 0 Points on ellipse: 1, 6, 3, 2

Since the major axis is horizontal, From the sketch, we can see that y h 1, k 2, a 4, b 2 (1, 6) x2 y2 6 1. a2 b2 x 12 y 22 1. 4 16 Substituting the values of the coordinates of the given (3, 2) (1, 2) points into this equation, we have x −4 −2 4 9 1 16 1, and 1. −2 a2 b2 a2 The solution to this system is a2 16, b2 167. Therefore, x2 y2 x2 7y2 1, 1. 16 167 16 16

y2 x2 x2 y2 45. 1 46. 1 1 4 25 9 a 1, b 2, c 5 a 5, b 3, c a2 b2 34 Center: 0, 0 y Center: 0, 0 y ± 4 ± 10 Vertices: 0, 1 Vertices: 5, 0 8 6 ± 2 ± Foci: 0, 5 Foci: 34, 0 4 2 1 x 3 x ± 4 2 2 4 ± − − Asymptotes: y x Asymptotes: y x 8 4 −2 4810 2 5 − 2 4 −6 −8 4 −10 366 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

x 12 y 22 y 12 x 42 47. 1 48. 1 4 1 122 52 a 2, b 1, c 5 a 12, b 5, c a2 b2 13 Center: 1, 2 Center: 4, 1 Vertices: 1, 2, 3, 2 Vertices: 4, 11, 4, 13 Foci: 1 ± 5, 2 Foci: 4, 14, 4, 12 1 12 Asymptotes: y 2 ± x 1 Asymptotes: y 1 ± x 4 2 5

y y

20 1

x 1 1 2 3

5 2 x 1 2 6 7 8 −5 4

5 − 20

49. 9 x2 y2 36x 6y 18 0 50. y2 9x2 36x 72 0 9x2 4x 4 y2 6y 9 18 36 9 y2 9x2 4x 4 72 36 36 x 22 y 32 y2 x 22 1 1 1 9 36 4 a 1, b 3, c 10 a 6, b 2, c a2 b2 210 Center: 2, 3 Center: 2, 0 Vertices: 1, 3, 3, 3 Vertices: 2, 6, 2, 6 Foci: 2 ± 10, 3 Foci: 2, 210, 2, 210 Asymptotes: y 3 ± 3x 2 Asymptotes: y ±3x 2

y y

x 2 2 4 6 5 2

x 4 −2 −154

−5 6

51. x2 9y2 2x 54y 80 0 52. 9x2 6x 9 4y2 2y 1 78 81 4 1 x2 2x 1 9y2 6y 9 80 1 81 0 9 x 32 4y 12 1 x 12 9y 32 0 y 12 x 32 1 14 19 1 y 3 ± x 1 3 1 1 13 a , b , c y 2 3 6 Degenerate hyperbola is two lines intersecting at 1, 3 . 3 Center: 3, 1 y 2 1 3 1 x Vertices: 3, , 3, 4 2 2 2 2 x 2 1 −3 −1 Foci: 3, 1 ± 13 6 −1 4 3 Asymptotes: y 1 ± x 3 6 2 Section 10.1 Conics and Calculus 367

53. 9 y2 x2 2x 54y 62 0 54. 9 x2 y2 54x 10y 55 0 9y2 6y 9 x2 2x 1 62 1 81 18 9x2 6x 9 y2 10y 25 55 81 25 y 32 x 12 1 1 2 18 x 32 y 52 1 a 2, b 32, c 25 1 19 1 −5 7 Center: 1, 3 1 10 a , b 1, c 10 Vertices: 1, 3 ± 2 3 3 Foci: 1, 3 ± 25 Center: 3, 5 −7 1 Solve for y: Vertices: 3 ± , 5 3 −8 2 9y2 6y 9 x2 2x 62 81 0 10 Foci: 3 ± , 5 x2 2x 19 3 y 32 9 Solve for y: 1 y 3 ± x2 2x 19 y2 10y 25 9x2 54x 55 25 3 y 52 9x2 54x 80 (Graph each curve separately.) y 5 ± 9x2 54x 80 (Graph each curve separately.)

55. 3 x2 2y2 6x 12y 27 0 56. 3 y2 x2 6x 12y 0 3x2 2x 1 2y2 6y 9 27 3 18 12 3y2 4y 4 x2 6x 9 0 12 9 3 x 12 y 32 y 22 x 32 1 1 4 6 1 3

a 2, b 6, c 10 1 a 1, b 3, c 2 6 −5 7 Center: 1, 3 Center: 3, 2

Vertices: 1, 3, 3, 3 Vertices: 3, 1, 3, 3 −4 10 Foci: 1 ± 10, 3 Foci: 3, 0, 3, 4 −7 −4 Solve for y: Solve for y: 2y2 6y 9 3x2 6x 27 18 3y2 4y 4 x2 6x 12 3x2 6x 9 x2 6x 12 y 32 y 22 2 3 3x2 2x 3 x2 6x 12 y 3 ± y 2 ± 2 3 (Graph each curve separately.) (Graph each curve separately.)

57. Vertices: ±1, 0 58. Vertices: 0, ±3 Asymptotes: y ±3x Asymptotes: y ±3x Horizontal transverse axis Vertical transverse axis Center: 0, 0 a 3 b b a a 1, ± ± ±3 ⇒ b 3 Slopes of asymptotes: ± ±3 a 1 b x2 y2 Therefore, 1. Thus,b 1 . Therefore, 1 9 y2 x2 1. 9 1 368 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

59. Vertices: 2, ±3 60. Vertices: 2, ±3 Point on graph: 0, 5 Foci: 2, ±5 Vertical transverse axis Vertical transverse axis Center: 2, 0 Center: 2, 0 a 3 a 3, c 5, b2 c2 a2 16 Therefore, the equation is of the form y2 x 22 Therefore, 1. 9 16 y2 x 22 1. 9 b2 Substituting the coordinates of the point 0, 5 , we have 25 4 9 1 or b2 . 9 b2 4 y2 x 22 Therefore, the equation is 1. 9 94

61. Center: 0, 0 62. Center: 0, 0 Vertex: 0, 2 Vertex: 3, 0 Focus: 0, 4 Focus: 5, 0 Vertical transverse axis Horizontal transverse axis 2 2 2 a 2, c 4, b2 c2 a2 12 a 3, c 5, b c a 16 2 2 y2 x2 x y Therefore, 1. Therefore, 1. 4 12 9 16

63. Vertices: 0, 2, 6, 2 64. Focus: 10, 0 2 2 3 Asymptotes: y x, y 4 x Asymptotes: y ± x 3 3 4 Horizontal transverse axis Horizontal transverse axis Center: 3, 2 Center:0, 0 since asymptotes intersect at the origin. a 3 c 10 b 2 b 3 3 Slopes of asymptotes: ± ± Slopes of asymptotes:± ± and b a a 3 a 4 4 Thus,b 2. Therefore, c2 a2 b2 100 x 32 y 22 Solving these equations, we have a2 64 and b2 36. 1. 9 4 Therefore, the equation is

x2 y2 1. 64 36

x2 2x x 65. (a) y2 1, 2yy 0, y (b) From part (a) we know that the slopes of the normal lines 9 9 9y must be 923 . ±6 ±23 At x 6: y ±3, y 9 93 9 At 6, 3 : y 3 x 6 23 23 At 6, 3 : y 3 x 6 or 9x 23y 60 0 9 9 or 2x 33y 3 0 At 6, 3 : y 3 x 6 23 23 At 6, 3 : y 3 x 6 or 9x 23y 60 0 9 or 2x 33y 3 0 Section 10.1 Conics and Calculus 369

y2 x2 66. (a) 1, y2 2x2 4, 2yy 4x 0, 67. x2 4y2 6x 16y 21 0 4 2 x2 6x 9 4 y2 4y 4 21 9 16 4x 2x y 2 2 2y y x 3 4y 2 4 ±24 4 Ellipse At x 4: y ±6, y ± 6 3 4 At 4, 6: y 6 x 4 or 4x 3y 34 0 3 4 At 4, 6: y 6 x 4 or 4x 3y 2 0 3 (b) From part (a) we know that the slopes of the normal lines must be 34. 3 At 4, 6:y 6 x 4 or 3x 4y 36 0 4 3 At 4, 6: y 6 x 4 or 3x 4y 36 0 4

68. 4 x2 y2 4x 3 0 69. y2 4y 4x 0

2 1 2 2 4 x x 4 y 3 1 y 4y 4 4x 4 12 2 2 4 x 2 y 4 y 2 4 x 1 Hyperbola Parabola

70. 25x2 10x 200y 119 0 71. 4 x2 4y2 16y 15 0

2 2 1 2 2 25 x 5x 25 200y 119 1 4 x 4 y 4y 4 15 16 12 2 2 25 x 5 200 y 1 4 x 4 y 2 1 Parabola Circle (Ellipse)

72. y2 4y x 5 73. 9 x2 9y2 36x 6y 34 0

2 2 2 2 1 y 4y 4 x 5 4 9 x 4x 4 9 y 3y 9 34 36 1 2 2 12 y 2 x 9 9 x 2 9 y 3 3 Parabola Circle (Ellipse)

74. 2 xx y y3 y 2x 75. 3 x 12 6 2 y 12 2 x2 2xy 3y y2 2xy 3 x 12 2y 1)2 6 2 2 2 x y 3y 0 x 12 y 12 1 2 3 3 2 9 2 x2 y 2 4 Hyperbola Ellipse

76. 9 x 32 36 4y 22 77. (a) A parabola is the set of all points x, y that are equidistant from a fixed line (directrix) and a fixed 9 x 32 4y 22 36 point (focus) not on the line. x 32 y 22 1 (b)x h2 4py k or y k2 4px h 4 9 (c) See Theorem 10.2. Ellipse 370 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

78. (a) An ellipse is the set of all points x, y , the sum of 79. (a) A hyperbola is the set of all points x, y for which the whose distance from two distinct fixed points (foci) absolute value of the difference between the distances is constant. from two distance fixed points (foci) is constant. x h2 y k2 x h2 y k2 x h2 y k2 y k2 x h2 (b) 1 or 1 (b) 1 or 1 a2 b2 b2 a2 a2 b2 a2 b2 b a (c)y k ± x h or y k ± x h a b

c 80. e , c a2 b2, 0 < e < 1 81. Assume that the vertex is at the origin. a 2 x 4py y For e 0, the ellipse is nearly circular. 2 3 4p1 3 For e 1, the ellipse is elongated. Focus 9 p 2 4 (−3, 1) (3, 1) 1 The pipe is located 9 x −3 −2 −1 1 2 3 4 meters from the vertex.

82. Assume that the vertex is at the origin. 83. y ax2 (a) x2 4py y 2ax 3 The equation of the tangent line is 8 2 4p 100 2 y ax0 2ax0 x x0 1600 p or y 2ax x ax 2. 3 0 0 Let y 0. Then: 1600 6400 2 x 4 y y 2 2 3 3 ax0 2ax0x 2ax0 2 (b) The deflection is 1 cm when ax0 2ax0x 2 128 x y ⇒ x ± ±6.53 meters. x 0 100 3 2

y x Therefore, 0, 0 is the x-intercept. 5 100 2

4 100 y − 3 3 3 ()8, 100 ()0, 100 ()8, 100

2 100

1 100

2 x 2 (,xax ) −8 −448 yax= 00 x x 0, 0 ()2 Section 10.1 Conics and Calculus 371

84. (a) Without loss of generality, place the coordinate system (b) x2 4x 4y 0 so that the equation of the parabola is x2 4py and, dy hence, 2x 4 4 0 dx 1 y x. dy 1 2p x 1 dx 2 Therefore, for distinct tangent lines, the slopes are At 0, 0, the slope is 1: y x. At 6, 3, the slope is unequal and the lines intersect. 2:y 2x 9. Solving for x, x 2x 9 3x 9 x 3 y 3. Point of intersection: 3, 3

2 2 85. (a) Consider the parabola x 4py. Let m0 be the slope of (b) x 4x 4y 8 0 the one tangent line at x1, y1 and therefore,1 m0 is x 22 4y 1 the slope of the second at x2, y2 . Differentiating, x 2x 4py or y , and we have: Vertex: 2, 1 2p dy 1 2x 4 4 0 m x or x 2pm dx 0 2p 1 1 0 dy 1 1 1 2p x 1 x2 or x 2 . dx 2 m0 2p m0

2 dy 5 dy 1 Substituting these values of x into the equation x At 2, 5, 2. At 3, , . 4py, we have the coordinates of the points of tangency dx 4 dx 2 2 2 2pm0, pm0 and 2p m0, p m0 and the equations Tangent line at 2, 5: of the tangent lines are y 5 2x 2 ⇒ 2x y 1 0. y pm 2 m x 2pm and 5 0 0 0 Tangent line at 3, : 4 p 1 2p y x . m 2 m m 5 1 0 0 0 y x 3 ⇒ 2x 4y 1 0. 4 2 The point of intersection of these lines is

2 1 p m0 1 Since m m 2 1, the lines are perpendicular. , p and is on the directrix,y p . 1 2 2 m0 1 1 Point of intersection: 2x 1 x y 2 4 xpy2 = 4 2p p − , 5 5 ()m m2 0 0 x 2 4

1 (2pm , pm2 ) x 00 2 x yp= − y 0 pm(1)2 − 0 , −p ()m 1 0 Directrix:y 0 and the point of intersection , 0 lies 2 on this line. 372 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

86. The focus of x2 8y 42y is 0, 2. The distance from a point on the parabola,x, x28, and the focus,0, 2 , is x2 2 d x 02 2 . 8 Since d is minimized when d 2 is minimized, it is sufficient to minimize the function y 4 2 2 2 x f x x 2 . 3 xy2 = 8 8 (0, 2) 2 x, x ()8 x2 x x3 1 fx 2x 2 2 x. x 8 4 16 −3 −2 −1 123

3 2 x x −2 fx 0 implies that x x 1 0 ⇒ x 0. 16 16 This is a minimum by the First Derivative Test. Hence, the closest point to the focus is the vertex, 0, 0.

87. y x x2 dy 1 2x dx y1 1 At the point of tangency x1, y1 on the mountain,m 1 2x1. Also, m . x1 1 y1 1 1 2x1 x1 1 2 x1 x1 1 1 2x1 x1 1 2 2 x1 x1 1 2x1 x1 1 2 x1 2x1 2 0 2 ± 22 412 2 ± 23 x 1 ± 3 1 21 2 Choosing the positive value for x1, we have x1 1 3. m 1 21 3 3 23 y 0 1 1 2 m x 1 x 1 − 1 0 0 ( 1, 1) (,xy ) 11(,x 0) 0 x 1 −2 −1 1 Thus, 3 2 3 x0 1 −1

1 −2 x 1 3 23 0 3 23 1 x 3 0 23 x . 3 0 The closest the receiver can be to the hill is 233 1 0.155. Section 10.1 Conics and Calculus 373

dA 88. (a) A 0.75t2 9.2t 301 (c) 1.5t 9.2, 6 ≤ t ≤ 12 dt (b) 350 y

14 12 10 8 513 6 200 4 2 x −2 2468101214

Women are spending more time watching TV each year.

89. Parabola Circle Vertex: 0, 4 Center: 0, k x2 4py 4 Radius: 8 y 2 2 42 4p0 4 x y k 64 2 2 p 1 4 0 k 64 x − − − 2 2 6 4 2 246 k 48 x 4y 4 −2 8 −4 x2 k 4 3 (Center is on the negative y-axis.) y 4 −6 2 2 4 −8 x y 4 3 64 y 43 ± 64 x2

Since the y- value is positive when x 0, we have y 43 64 x2. 4 x2 A 2 4 43 64 x2 dx 0 4 x3 1 x 4 24x 43x x64 x2 64 arcsin 12 2 8 0 64 1 216 163 248 32 arcsin 12 2 164 33 2 15.536 square feet 3

1 90. x y2 4 1 x y 2 y2 1 x2 1 4 4 y2 1 4 s 1 dy 4 y2 dy 0 4 2 0 1 4 y4 y2 4 lny 4 y2 4 0 1 420 4 ln 4 20 4 ln 2 4 25 ln2 5 5.916 374 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

91. (a) Assume that y ax2. y (−60, 20) (60, 20) 2 1 1 20 20 a602 ⇒ a ⇒ y x2 360 180 180 15 1 1 (b) f x x2, fx x 10 180 90 5 60 1 2 2 60 S 2 1 x dx 902 x2 dx x −60 −45 −30 −15 15 30 45 60 0 90 90 0 2 1 60 x902 x2 902 lnx 902 x2 (Formula 26) 90 2 0 1 6011,700 902 ln60 11,700 902 ln 90 90 1 180013 902 ln60 3013 902 ln 90 90 60 3013 2013 90 ln 90 2 13 10213 9 ln 128.4 m 3

1 1 3 92. x2 20y 93. x2 4py, p , , 1, , 2 4 2 2 x2 y As p increases, the graph becomes wider. 20 1 y p = p = 1 x 4 y 10 p = 2 24 3 p = r 2 r 2 2 x x 100 x S 2 x1 dx 2 dx 1 p = 0 10 0 10 2 2 r 232 232 x 100 x 100 r 1000 −16 −8 8 16 10 3 0 15

h 94. A 2 4py dy 95. (a) At the vertices we notice that 0 the string is horizontal and h has a length of 2a. 4p y12 dy 0 (b) The thumbtacks are located 2 h at the foci and the length of Focus Focus 32 4 p y string is the constant sum of Vertex Vertex 3 0 the distances from the foci. 8 ph32 3

5 5 2 3 96. a , b 2, c 22 97. y 2 2 2 17 16 15 14 13 12 11 10 9 8 The tacks should be placed 1.5 feet from the center. The 7 6 5 4 3 2 1 1 x 2 string should be 2a 5 feet long. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Section 10.1 Conics and Calculus 375

c c 98. e 99. e a a c A P 2a 0.0167 149,598,000 A P a c 2,498,286.6 2 Least distance: a c 147,099,713.4 km A P A P c a P P 2 2 Greatest distance: a c 152,096,286.6 km c A P2 A P e a A P2 A P

c ( a , 0) x A P

A P 100. e A P 35.29 0.59 A P 101. e 0.9671 A P 35.29 0.59 123,000 4000 119 4000 123,000 4000 119 4000 122,881 0.9372 131,119

x2 y2 x2 y2 102. 1 103. 1 a2 b2 102 52 x2 y2 2x 2yy 1 0 a2 a2b2a2 102 52 x2 y2 52x x 1 y a2 a2a2 c2a2 102y 4y x2 y2 8 2 1 At 8, 3: y a2 a21 e2 12 3 → 2 → 2 As e 0, 1 e 1 and we have The equation of the tangent line is y 3 3 x 8 . It 2 25 x2 y2 will cross the y -axis when x 0 and y 3 8 3 3 . 1 or the circle x2 y2 a2. a2 a2

x2 y2 104. 1 y 4.52 2.52 y2 x2 4.52 1 2 2.5 5 ft x 9 3 ft x ± 2.52 y2 5 9 ft V Area of bottomLength Area of topLength 4.52.5 0.5 9 V 16 16 2 2.52 y2 dy(Recall: Area of ellipse is ab .) 2 0 5 144 y 0.5 144 1 90 y2.52 y2 2.52 arcsin 90 0.56 2.52 arcsin 354.3 ft3 5 2.5 0 5 5 376 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

105. 16x2 9y2 96x 36y 36 0 106. 9x2 4y2 36x 24y 36 0 32x 18yy 96 36y 0 18x 8yy 36 24y 0 y18y 36 32x 96 8y 24y 18x 36 32x 96 18x 36 y y 18y 36 8y 24 y 0 when x 3. y is undefined when y 2. y 0 when x 2. y undefined when y 3. At x 2, y 0 or 6. At x 3, y 2 or 6. Endpoints of major axis: 2, 0, 2, 6 Endpoints of major axis: 3, 2, 3, 6 At y 3, x 0 or 4. At y 2, x 0 or 6. Endpoints of minor axis: 0, 3, 4, 3 Endpoints of minor axis: 0, 2, 6, 2 x 22 y 32 x 32 y 22 Note: Equation of ellipse is 1 Note: Equation of ellipse is 1 4 9 9 16

2 1 x 2 107. (a) A 4 4 x2 dx x4 x2 4 arcsin 2 or, A ab 21 2 0 2 2 0 2 1 1 1 2 8 (b) Disk: V 2 4 x2 dx 4x x3 0 4 2 3 0 3 1 y 4 x2 2 x y 24 x2 x2 16 3x2 1 y2 1 16 4x2 4y 2 16 3x2 2 S 22 y dx 16 3x2 dx 0 4y 0 3x 2 3x16 3x2 16 arcsin 23 4 0 2 9 43 21.48 9 2 2 2 2 16 (c) Shell: V 2 x4 x2 dx 2x4 x212 dx 4 x232 0 0 3 0 3 x 21 y2 2y x 1 y2 4y2 1 3y2 1 x2 1 1 y2 1 y2 1 1 3y2 1 S 22 21 y2 dy 8 1 3y2 dy 2 0 1 y 0 8 1 3y1 3y2 ln3y 1 3y2 23 0 4 6 3 ln2 3 34.69 3 Section 10.1 Conics and Calculus 377

4 3 3 x 4 108. (a) A 4 16 x2 dx x16 x2 16 arcsin 12 0 4 2 4 0 4 9 9 1 4 (b) Disk: V 2 16 x2 dx 16x x3 48 0 16 8 3 0 3 y 16 x2 4 3x y 416 x2 9x2 1 y2 1 1616 x2 4 3 1616 x2 9x2 4 3 256 7x2 3 4 S 22 16 x2 dx 4 16 x2 dx 256 7x2 dx 2 2 0 4 16 16 x 0 4 416 x 4 0 3 7x 4 3 7 7x256 7x2 256 arcsin 487 256 arcsin 138.93 87 16 0 87 4 4 3 1 2 4 (c) Shell: V 4 x 16 x2 dx 3 16 x232 64 0 4 2 3 0 4 x 9 y2 3 4y x 39 y2 16y2 1 x2 1 99 y2 3 4 99 y2 16y2 S 22 9 y2 dy 2 0 3 9 9 y 3 4 4 81 7y2 dy 0 9 16 3 7y81 7y2 81 ln7y 81 7y2 9 27 0 8 3712 81 ln37 12 81 ln 9 168.53 97

109. From Example 5,

2 C 4a 1 e2 sin2 d 0 x2 y2 For 1, we have 25 49 c 26 a 7, b 5, c 49 25 26, e . a 7 2 24 C 47 1 sin2 d 0 49 281.3558 37.9614 378 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

x2 y2 y 110. (a) 1 (b) Slope of line through c, 0 and x , y : m 0 2 2 0 0 1 a b x0 c 2x 2yy y 0 Slope of line through c, 0 and x , y : m 0 2 2 0 0 2 a b x0 c xb2 y ya2

2 b x0 At P, y 2 m. a y0

2 y0 b x0 m m x c a2y a2y 2 b2x x c (c) tan 2 0 0 0 0 0 1 m m y b2x a2y x c b2x y 2 1 0 0 0 0 0 0 2 x0 c a y0 a2y 2 b2x 2 b2x c a2b2 b2x c b2a2 x c b2 0 0 0 0 0 2 2 2 2 2 2 x0y0 a b a y0c x0y0c a y0c y0c x0c a y0c b2 b2 arctan arctan y0c y0c

2 y0 b x0 m m x c a2y a2y 2 b2x x c tan 1 0 0 0 0 0 1 m m y b2x a2y x c b2x y 1 1 0 0 0 0 0 0 2 x0 c a y0 a2y 2 b2x 2 b2x c a2b2 b2x c b2a2 x c b2 0 0 0 0 0 2 2 2 2 2 2 2 a x0 y0 a cy0 b x0y0 x0 y0 a b a cy0 y0c x0c a y0c b2 arctan y0c Since , the tangent line to an ellipse at a point P makes equal angles with the lines through P and the foci.

c a2 b2 111. Area circle r2 100 112. (a)e ⇒ ea2 a2 b2. Hence, a a Area ellipse ab a10 x h2 y k2 1 2100 10a ⇒ a 20 a2 b2 Hence, the length of the major axis is 2a 40. x h2 y k2 1. a2 a21 e2 x 22 y 32 (b) 1 4 41 e2

−39

−1

113. The transverse axis is horizontal since 2, 2 and 10, 2 are the foci (see definition of hyperbola). Center: 6, 2 c 4, 2a 6, b2 c2 a2 7 Therefore, the equation is x 62 y 22 1. 9 7 Section 10.1 Conics and Calculus 379

114. The transverse axis is vertical since 3, 0 and 3, 3 115. 2a 10 ⇒ a 5 are the foci. ⇒ c 6 b 11 16 17 14 15 13 11 12 10 3 8 9 6 7 4 5 Center: 3, 3 1 2 2 2 1 4 3 5 6 8 7 10 9 11 13 12 3 5 15 14 c , 2a 2, b2 c2 a2 17 16 2 4 Therefore, the equation is y 322 x 32 1. 1 54

116. Center: 0, 0 Horizontal transverse axis Foci: ±c, 0 Vertices: ±a, 0 The difference of the distances from any point on the hyperbola is constant. At a vertex, this constant difference is a c c a 2a. Now, for any point x, y on the hyperbola, the difference of the distances between x, y and the two foci must also be 2a. x c2 y 02 x c2 y 02 2a x c2 y2 2a x c2 y2 x c2 y2 4a2 4ax c2 y2 x c2 y2 4xc 4a2 4ax c2 y2

xc a2 ax c2 y2 y

2 2 2 4 2 2 2 2 x c 2a cx a a x 2cx c y (,x y ) 2 2 2 2 2 2 2 2 x c a a y a c a (,c 0) x x2 y2 (,−c 0) (,−a 0) ( a , 0) 1 a2 c2 a2 Since a2 b2 c2, we have x2a2 y2b2 1.

2c x c2 y2 117. Time for sound of bullet hitting target to reach x, y: y vm vs (,x y ) x c2 y2 Time for sound of rifle to reach x, y: vs x 2c x c2 y2 x c2 y2 (,−c 0) (,c 0) Since the times are the same, we have: rifle target vm vs vs 4c2 4c x c2 y2 x c2 y2 2 2 2 x c y 2 2 vm vmvs vs vs v 2x v 2c x c2 y2 m s vsvm v 2 v 2 m 2 2 s 2 1 2 x y 2 1c vs vm x2 y2 1 2 2 2 2 2 2 2 c vs vm c vm vs vm 380 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

118. c 150, 2a 0.001186,000, a 93, b 1502 932 13,851 x2 y2 1 932 13,851 When y 75, we have 752 x2 9321 13,851 x 110.3 miles.

x2 y2 119. The point x, y lies on the line between 0, 10 and 10, 0. 120. 1 a2 b2 Thus,y 10 x. The point also lies on the hyperbola x236 y264 1. Using substitution, we have: 2x 2yy b2x 0 or y a2 b2 a2y x2 10 x2 1 36 64 b2x y y 0x x 0 a2y 0 16x2 910 x2 576 0 a2y y a2y 2 b2x x b2x 2 7x2 180x 1476 0 0 0 0 0 b2x 2 a2y 2 b2x x a2y y 180 ± 1802 471476 0 0 0 0 x 2 2 2 2 2 7 a b b x0x a y0y 180 ± 1922 90 ± 962 x x y y 0 0 1 14 7 a2 b2 Choosing the positive value for x we have: 90 962 x 6.538 and 7 160 962 y 3.462 7

x2 2y2 2y 2 x2 121. 1 ⇒ 1 . Let c2 a2 b2. a2 b2 b2 a2 x2 2y2 2y2 x2 1 ⇒ 1 a2 b2 b2 b2 a2 b2 x2 x2 1 1 1 1 ⇒ 2 x2 a2 a2 b2 a2 a2 b2 2a2a2 b2 2aa2 b2 2ac x2 ⇒ x ± ± 2a2 b2 2a2 b2 2a2 b2 2y2 1 2a2c2 2y2 b2 1 ⇒ b2 a2 2a2 b2 b2 2a2 b2 b4 b2 y2 ⇒ y ± 22a2 b2 22a2 b2 2ac b2 2ac b2 There are four points of intersection: , ± , , ± 2a2 b2 22a2 b2 2a2 b2 22a2 b2 x2 2y2 2x 4yy b2x 1 ⇒ 0 ⇒ y a2 b2 a2 b2 e 2a2y x2 2y2 2x 4yy b2x 1 ⇒ 0 ⇒ y a2 b2 b2 c2 b2 h 2c2y —CONTINUED— Section 10.1 Conics and Calculus 381

121. —CONTINUED— 2ac b2 At , , the slopes of the tangent lines are: 2a2 b2 22a2 b2 2ac 2ac b2 b2 2 2 c 2 2 a y 2a b and y 2a b . e b2 a h b2 c 2a2 2c2 22a2 b2 22a2 b2 Since the slopes are negative reciprocals, the tangent lines are perpendicular. Similarly, the curves are perpendicular at the other three points of intersection.

122. Ax2 Cy2 Dx Ey F 0 (Assume A 0 and C 0; see (b) below) D E Ax2 x Cy2 y F A C D D2 E E2 D2 E2 Ax2 x Cy2 y F R A 4A2 C 4C2 4A 4C D 2 E 2 x y 2A 2C R C A AC (a) If A C, we have (b) If C 0, we have D 2 E 2 R D 2 D 2 x y Ax F Ey . 2A 2C A 2A 4A which is the standard equation of a circle. If A 0, we have

(c) If AC < 0, we have E 2 E2 Cy F Dx . 2C 4C D 2 E 2 x y 2A 2C These are the equations of parabolas. 1 R R (d) If AC < 0, we have A C D 2 E 2 which is the equation of an ellipse. x y 2A 2C ±1 R R A C which is the equation of a hyperbola.

123. False. See the definition of a 124. True 125. True parabola.

126. False. y2 x2 2x 2y 0 127. True 128. True yields two intersecting lines: y 1 ± x 1

x2 y2 129. Let 1 be the equation of the ellipse with a > b > 0. Let ±c, 0 be the foci, a2 b2 c2 a2 b2. Let u, v be a point on the tangent line at Px, y, as indicated in the figure.

x2b2 y2a2 a2b2 y 2 xb2 2yya2 0 b (u, v) −a d P(x, y) 2 b x x y Slope at Px, y −c c 2 F1 F2 a y a

−b x2 y2 + = 1 2 2 —CONTINUED— a b 382 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

129. —CONTINUED— y v b2x Now, x u a2y y2a2 a2vy b2x2 b2xu y2a2 x2b2 a2vy b2ux a2b2 a2vy b2ux Since there is a right angle at u, v, v a2y u b2x vb2x a2uy. We have two equations: a2vy b2ux a2b2 a2uy b2vx 0. Multiplying the first by v and the second by u, and adding, a2v2y a2u2y a2b2v yu2 v2 b2v yd2 b2v yd2 v . b2 xd2 Similarly, u . a2 xd yd From the figure,u d cos and v d sin . Thus,cos and sin . a2 b2 x2d2 y2d2 cos2 sin2 1 a4 b4 x2b4d2 y2a4d2 a4b4 a4b4 d2 x2b4 y2a4 Let r1 PF1 and r2 PF2, r1 r2 2a. 1 r r r r 2 r 2 r 2 1 2 2 1 2 1 2 1 4a2 x c2 y2 x c2 y2 2 2a2 x2 y2 c2 a2 b2 x2 y2 a4b4 Finally, d2r r a2 b2 x2 y2 1 2 x2b4 y2a4 a4b4 a2 b2 x2 y2 b2b2x2 a2a2y2 a4b4 a2 b2 x2 y2 b2a2b2 a2y2 a2a2b2 b2x2 a4b4 a2 b2 x2 y2 a2b2a2 b2 x2 y2 a2b2, a constant! Section 10.2 Plane Curves and Parametric Equations 383

y 2 2 9 9 130. Consider circle x y 2 and hyperbola y . y = x x 5 4 9 v, 9 3 ()v Let u, 2 u2 and v, be points on the circle and v 2 hyperbola, respectively. We need to minimize the distance ()u, 2 − u2 x between these 2 points: −2 2 3 4 5

− x2 + y2 = 2 9 2 2 Distance2 f u, v u v2 2 u2 . v The tangent lines at 1, 1 and 3, 3 are both perpendicular to y x, and hence parallel. The minimum value is 3 12 3 12 8.

Section 10.2 Plane Curves and Parametric Equations

1. x t, y 1 t

(a) (b) y t 01 2 3 4 1 x 012 3 2 x −1 1 2 3 y 1 0 1 2 3 −1

−2

−1 3 (d) x2 t y 1 x2, x ≥ 0 −4

2. x 4 cos2 y 2 sin 0 ≤ x ≤ 4 2 ≤ y ≤ 2

(a) (c) 3 0 2 4 4 2 −1 5 x 02420

y 2 2 022 −3

x (b) y (d) cos2 4 3 2 2 y sin2 1 4 x 123 5 x y2 1 −2 4 4 x 4 y2, 2 ≤ y ≤ 2 (e) The graph would be oriented in the opposite direction.

3. x 3t 1 y 4. x 3 2t y

4 y 2t 1 y 2 3t

6 x 1 2 3 x y 2 1 y 2 3 3 2 4 x 2x 3y 5 0 42 2y 3x 13 0 2 2 x 2468 384 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

5. x t 1 6. x 2t2 7. x t3

2 4 1 2 y t y t 1 y 2t y x 12 x 2 x2 x t3 implies t x13 y 1 1, x ≥ 0 y 2 4 1 23 y 2x

For t < 0, the orientation is right y 4 to left.

For t > 0, the orientation is left to 1

right. x 3 2 1 1 2 3 x y 2 2 4

6 5 4 3 2 1 x −1 123456 −1

8. x t2 t, y t2 t Subtracting the second equation from the first, we have x y x y 2t or t . t 2 1 0 1 2 y 2 x 20026 4 x y2 x y 3 y y 62002 4 2 2 Since the discriminant is x B2 4AC 22 411 0, −1234 −1 the graph is a rotated parabola.

9. x t, t ≥ 0 10. x 4 t, t ≥ 0 11. x t 1 y t 2 y 3 t t y t 1 y x2 2, x ≥ 0 y 3 x4, x ≥ 0 x 1 y y y x 4 3

3 2 y 2 1 4 1 x −3 −2 −1 123 x −1 2 3 456 2 −1 −2 −2 − 3 x 2 2

1 12. x 1 y 13. x 2t y t 8 1 y t 2 y t 1 x 4 −2 2 x x 4 1 1 y 2 x 1 implies t 2 2 x t x 1 448 12 1 y 1 −3 x 1 Section 10.2 Plane Curves and Parametric Equations 385

14. x t 1 15. x et, x > 0 16. x et, x > 0 y t 2 y e3t 1 y e2t 1 y x3 1, x 0 1 x y 2 1 y 3 > 2 y x 1 2 1, x > 0 y y x

y 5 5 4 4 3 3 2 3 2 1 2 1 x − − − 1 3 2 1 13 x −1 −2 −1 1 2 34 − x 1 −2 1 2453 −3

17. x sec 18. x tan2 19. x 3 cos , y 3 sin y cos y sec2 Squaring both equations and adding, we have sec2 tan2 1 0 ≤ < , < ≤ 2 2 x2 y2 9. y x 1 y xy 1 ≥ x 0 4 1 y y x 2 4 x x ≥ 1, y ≤ 1 42 2 4 3 2 y 2 3 4 1 2

1 x 1 243 x 123

20. x 2 cos 21. x 4 sin 2 22. x cos y 6 sin y 2 cos 2 y 2 sin 2 x 2 y 2 x2 y 4 sin cos cos2 sin2 1 sin2 2 2 6 16 1 x2 sin2 x2 y2 y2 1 ellipse cos2 2 y ±4x1 x2 4 36 4 3 y x2 y2 1 16 4 −22 4 4 2

x −6 −4 46 −3 −2 −6 6

−4 386 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

23. x 4 2 cos 24. x 4 2 cos y 1 sin y 1 2 sin 2 x 4 2 2 cos2 x 4 4 cos 4 y 12 4 sin2 y 12 sin2 1 x 42 y 12 4 x 42 y 12 3 1 4 1 −17 2

−1 8 −5

−4

25. x 4 2 cos 26. x sec 27. x 4 sec y 1 4 sin y tan y 3 tan x 42 x2 sec2 x2 cos2 sec2 4 16 y2 tan2 y 12 y2 sin2 x2 y2 1 tan2 16 9 2 x 42 y 12 x2 y2 1 1 4 16 16 9 −33 3 6

−2 −2 10 −9 9

−5 −6

28. x cos3 29. x t3 30. x ln 2t y sin3 y 3 ln t y t2 x23 cos2 y 3 ln 3 x ln x ex t 2 y2 3 sin2 2 e2x 1 x23 y23 1 y e2x −1 5 r 4 1.5 3

−2 −22

−23

−1.5 −1 Section 10.2 Plane Curves and Parametric Equations 387

2t 31. x e t 3 32. x e 4 y e3t y et 1 y2 x et −1 5 −13 x − −1 y > 0 1 et 3 y y x, x > 0 1 3 y x 1 y x3 x > 0 y > 0

33. By eliminating the parameters in (a) Ð (d), we get y 2x 1. They differ from each other in orientation and in restricted domains. These curves are all smooth except for (b). (a) x t, y 2t 1 (b) x cos y 2 cos 1

y 1 ≤ x ≤ 1 1 ≤ y ≤ 3 3 dx dy 0 when 0, ±, ±2, . . . . 2 d d

1 y

x 3 −2 −1 12 − 1 2

x −2 −1 12

−1

y y

4 4

3 3

2 2

1 1

x x −1 132 −1 132 388 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

34. By eliminating the parameters in (a) Ð (d), we get x2 y2 4. They differ from each other in orientation and in restricted domains. These curves are all smooth. 4t2 1 1 1 (a) x 2 cos , y 2 sin (b) x 4 y t t2 t y x ≥ 0, x 2 y 0 3 y

1 2 x −3 −1 13 1

x −1 13 −3 −1

−2

y y

3 3

1 1

x x 123 −3 −2 −1

35. The curves are identical on 0 < < . They are both smooth. Represent y 21 x2

36. The orientations are reversed. The graphs are the same. They are both smooth.

37. (a) 4 4 (b) The orientation of the second curve is reversed. (c) The orientation will be reversed. −6 6 −6 6 (d) Many answers possible. For example, x 1 t, y 1 2t, and x 1 t, x 1 2t. −4 −4

38. The set of points x, y corresponding to the rectangular equation of a set of parametric equations does not show the orientation of the curve nor any restriction on the domain of the original parametric equations.

39. x x1 t x2 x1 40. x h r cos y y1 t y2 y1 y k r sin x x1 x h t cos x2 x1 r x x1 y k y y1 y2 y1 sin x2 x1 r y y x h2 y k2 y y 2 1x x cos2 sin2 1 1 1 2 2 x2 x1 r r 2 2 2 y y1 m x x1 x h y k r Section 10.2 Plane Curves and Parametric Equations 389

41. x h a cos 42. x h a sec y k b sin y k b tan x h x h cos sec a a y k y k sin tan b b x h2 y k2 x h2 y k2 1 1 a2 b2 a2 b2

43. From Exercise 39 we have 44. From Exercise 39 we have 45. From Exercise 40 we have x 5t x 1 4t x 2 4 cos y 2t. y 4 6t. y 1 4 sin . Solution not unique Solution not unique Solution not unique

46. From Exercise 40 we have 47. From Exercise 41 we have 48. From Exercise 41 we have x 3 3 cos a 5, c 4 ⇒ b 3 a 5, c 3 ⇒ b 4 y 1 3 sin . x 5 cos x 4 5 cos Solution not unique y 3 sin . y 2 4 sin . Center: 0, 0 Center: 4, 2 Solution not unique Solution not unique

49. From Exercise 42 we have 50. From Exercise 42 we have 51. y 3x 2 a 4, c 5 ⇒ b 3 a 1, c 2 ⇒ b 3 Example x 4 sec x 3 tan x t, y 3t 2 y 3 tan . y sec . x t 3, y 3t 11 Center: 0, 0 Center: 0, 0 Solution not unique Solution not unique The transverse axis is vertical, therefore,x and y are interchanged.

2 52. y 53. y x3 54. y x2 x 1 Example Example Example x t, y t3 x t, y t 2 2 x t, y t 1 x 3 t, y t x t3, y t6 2 x tan t, y tan3 t x t, y t 1

3 55. x 2 sin 56. x sin 57. x 2 sin 3 y 2 1 cos y 1 cos y 1 2 cos

5 6 5

−66 −216 −27

− −1 2 −1 Not smooth at 2n Not smooth at x 2n 1 390 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

58. x 2 4 sin 59. x 3 cos3 60. x 2 sin y 2 4 cos y 3 sin3 y 2 cos

9 4 4

−66

−99

− 5 −3 −4 0 Not smooth at x, y ±3, 0 and Smooth everywhere ± 1 0, 3 , or 2n .

3t 61. x 2 cot 4 62. x 2 1 t3 y 2 sin2 3t2 −66y −33 1 t3

− −4 2

Smooth everywhere Smooth everywhere

63. See definition on page 709.

64. Each point x, y in the plane is determined by the plane curve x f t, y gt. For each t, plot x, y. As t increases, the curve is traced out in a specific direction called the orientation of the curve.

65. A plane curve C,represented by x f t, y gt , is smooth if f and g are continuous and not simultaneously 0. See page 714.

66. (a) Matches (iv) because 0, 2 is on the graph. (b) Matches (v) because 1, 0 is on the graph. (c) Matches (ii) because 1 ≤ x ≤ 0 and 1 ≤ y ≤ 3. (d) Matches (iii) because 4, 0 is on the graph. (e) Matches (vi) because undefined at 0. (f) Matches (i) because x y 22 1 for all y.

67. When the circle has rolled radians, we know that the center is at a, a. y C BD sin sin180 or BD b sin b b P b AP A C cos cos180 or AP b cos θ b a x BD Therefore,x a b sin and y a b cos .

68. Let the circle of radius 1 be centered at C. A is the point of tangency on the line OC. y OA 2, AC 1, OC 3. P x, y is the point on the curve being traced out as the 3 angle changes AB AP, AB 2 and AP ⇒ 2. Form the right triangle C CDP. The angle OCE 2 and 2 α D A DCP 3 . 1 Px= ( , y ) 2 2 2 θ x 1 BE x x OE Ex 3 sin sin3 3 cos cos 3 2 2 y EC CD 3 sin cos3 3 sin sin 3 2 Hence, x 3 cos cos 3, y 3 sin sin 3. Section 10.2 Plane Curves and Parametric Equations 391

69. False 70. False. Let x t2 and y t. Then x y2 and y is not a function of x. x t2 ⇒ x ≥ 0 y t2 ⇒ y ≥ 0 The graph of the parametric equations is only a portion of the line y x.

1005280 440 71. (a) 100 mihr ftsec (d) We need to find the angle (and time t ) such that 3600 3 440 440 x cos t 400 x v cos t cos t 3 0 3 440 y h v sin t 16t2 y 3 sin t 16t2 10. 0 3 440 3 sin t 16t2 From the first equation t 1200440 cos . Substituting 3 into the second equation, (b) 30 440 1200 1200 2 10 3 sin 16 3 440 cos 440 cos 120 2 7 400 tan 16 sec2 0 400 44 0 120 2 400 tan 16 tan2 1. It is not a home run—when x 400, y < 10. 44

60 (c) We now solve the quadratic for tan : 120 2 120 2 16 tan2 400 tan 7 16 0 44 44

0 400 ⇒ 0 tan 0.35185 19.4

Yes, it’s a home run when x 400, y > 10.

72. (a) x v0 cos t 2 y h v0 sin t 16t

2 x ⇒ x x t y h v0 sin 16 v0 cos v0 cos v0 cos 16 sec2 2 y h tan x 2 x v0 16 sec2 2 2 80 (b) y 5 x 0.005x h tan x 2 x (c) v0 h 5, tan 1 ⇒ , and 4

0 250 16 sec24 16 −5 0.005 2 2 2 v0 v0 (d) Maximum height: y 55 at x 100 32 v 2 6400 ⇒ v 80. 0 0.005 0 Range: 204.88 Hence, x 80 cos45t y 5 80 sin45t 16t2. 392 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

Section 10.3 Parametric Equations and Calculus

dy dydt 4 2 dy dydt 1 1. 2. 3t23 dx dxdt 2t t dx dxdt 13t23

dy dyd 2 cos sin dy dyd 12e2 1 1 3. 1 4. e32 dx dxd 2 sin cos dx dxd 2e 4 4e32 dy Note: x y 1 ⇒ y 1 x and 1 d

5. x 2t, y 3t 1 6. x t, y 3t 1 dy dydt 3 dy 3 6t 6 when t 1. dx dxdt 2 dx 12t d 2y d 2y 3t 0 Line 6 concave upwards dx2 dx2 12t

7. x t 1, y t2 3t 8. x t2 3t 2, y 2t dy 2t 3 dy 2 2 1 when t 1. when t 0. dx 1 dx 2t 3 3 d 2y d 2y 222t 32 4 4 2 concave upwards when t 0. dx2 dx2 2t 3 2t 33 27 concave downward

9. x 2 cos , y 2 sin 10. x cos , y 3 sin dy 2 cos dy 3 cos dy cot 1 when . 3 cot is undefined when 0. dx 2 sin 4 dx sin dx d 2y csc2 csc3 d 2y 3 csc2 3 d 2y 2 when . is undefined when 0. dx2 2 sin 2 4 dx2 sin sin3 dx2 concave downward

11. x 2 sec , y 1 2 tan 12. x t, y t 1 2 dy 2 sec dy 1 2 t 1 dx sec tan dx 12t 2 sec t 2 csc 4 when . 2 when t 2. tan 6 t 1 dy d 2y t 12t t12t 1 t 1 d 2 2 d y dx 2 csc cot dx 12t dx2 dx sec tan 1 d 1 when t 2. t 132 2 cot3 63 when . concave downward 6 concave downward Section 10.3 Parametric Equations and Calculus 393

13. x cos3 , y sin3 14. x sin , y 1 cos dy 3 sin2 cos dy sin 0 when . dx 3 cos2 sin dx 1 cos 1 cos cos sin2 tan 1 when . 2 2 4 d y 1 cos 2 2 2 dx 1 cos d y sec 1 dx2 3 cos2 sin 3 cos4 sin 1 1 when . 1 cos 2 4 sec4 csc 42 when . 3 3 4 concave downward concave upward

15. x 2 cot , y 2 sin2 16. x 2 3 cos , y 3 2 sin dy 4 sin cos dy 2 cos 2 2 sin3 cos cot dx 2 csc2 dx 3 sin 3 2 3 2 dy 33 dy At , , , and . At 1, 3, 0, and is undefined. 3 2 3 dx 8 dx 3 33 2 Tangent line: x 1 Tangent line: y x 2 8 3 dy At 2, 5, , and 0. 33x 8y 18 0 2 dx dy Tangent line: y 5 At 0, 2, , and 0. 2 dx 4 33 7 dy 23 At , 2 , , and . Tangent line: y 2 0 2 6 dx 3 1 dy 3 Tangent line: At 23, , , and . 2 6 dx 8 23 4 33 y 2 x 1 3 3 2 Tangent line: y x 23 2 8 23x 3y 43 3 0 3x 8y 10 0

1 17. x 2t, y t2 1, t 2 18. x t 1, y 1, t 1 t (a) 10 (a) 4

−3 5 −6 6

−4 −4 (b) At t 2, x, y 4, 3, and (b) At t 1, x, y 0, 2, and dx dy dy 2, 4, 2. dx dy dy dt dt dx 1, 1, 1. dt dt dx dy (c) 2. At 4, 3, y 3 2x 4 dy dx (c)1. At 0, 2, y 2 1x 0 dx y 2x 5. y x 2. (d) 10 (d) 4

(4, 3)

− 5 −5 5 3

−4 −4 394 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

3 19. x t2 t 2, y t3 3t, t 1 20. x 4 cos , y 3 sin , 4 (a) 5 (a) 4

−1 8 −6 6

−3 −4 (b) At t 1, x, y 4, 2, and 3 4 3 (b) At , x, y , , and dx dy dy 4 3, 0, 0. 2 2 dt dt dx dx dy 32 dy 3 22, , . dy dt dt 2 dx 4 (c) 0. At 4, 2, y 2 0x 4 dx dy 3 4 3 3 3 4 (c) . At , , y x y 2 dx 4 2 2 2 4 2 (d) 5 3 y x 32. (4, 2) 4

4 −1 8 (d)

−3 −6 6

−4

21. x 2 sin 2t, y 3 sin t crosses itself at the origin, 22. x 2 cos t, y 2t sin t crosses itself at a point x, y 0, 0. on the x -axis:2, 0. The corresponding t -values are t ±2. At this point,t 0 or t . dy dx dy 2 cos t dy 3 cos t 2 cos t, sin t, dt dt dx sin t dx 4 cos 2t dy 2 dy 3 3 At t : . At t 0: and y x. Tangent Line 2 dx dx 4 4 2 dy 3 3 Tangent line: y 0 x 2 At t , and y x. Tangent Line dx 4 4 2 4 y x

dy 2 At t : . 2 dx

2 Tangent line: y 0 x 2

2 4 y x

23. x t2 t, y t3 3t 1 crosses itself at the point x, y 2, 1. At this point,t 1 or t 2.

2 dy 3t 3 dx 2t 1 dy At t 1, 0 and y 1. Tangent Line dx dy 9 At t 2, 3 and y 1 3x 2 or y 3x 5. Tangent Line dt 3 Section 10.3 Parametric Equations and Calculus 395

24. x t3 6t, y t2 crosses itself at 0, 6. The corresponding t -values are t ±6.

dy 2t dx 3t2 6 dy 26 6 At t 6, . dx 12 6 6 Tangent line: y 6 x 0 6 6 y x 6 6 dy 26 6 At t 6, . dx 12 6 6 Tangent line: y x 6 6

25. x cos sin , y sin cos dy Horizontal tangents: sin 0 when ±, ±2, ±3, . . . d Points:1, 2n 1, 1, 2n where n is an integer. Points shown: 1, 0, 1, , 1, 2 dx 3 5 Vertical tangents: cos 0 when ± , ± , ± , . . . d 2 2 2 Note: 0 corresponds to the cusp at x, y 1, 0. dy sin tan 0 at 0 dx cos 1n12n 1 3 5 Points: , 1n1 Points shown: , 1, , 1, , 1 2 2 2 2

26. x 2, y 21 cos 27. x 1 t, y t2 dy dy Horizontal tangents: 2 sin 0 when 0, ±, Horizontal tangents: 2t 0 when t 0 d dt ±2, . . . Point: 1, 0 Points:4n, 0, 22n 1, 4 where n is an integer dx Vertical tangents:1 0; none dt Points shown: 0, 0, 2, 4, 4, 0 3 dx Vertical tangents: 2 0; none d

−2 4 (1, 0)

−1

28. x t 1, y t2 3t dy 3 Horizontal tangents: 2t 3 0 when t dt 2 1 9 Point: , 2 4 dx Vertical tangents: 1 0; none dt 396 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

29. x 1 t, y t3 3t 30. x t2 t 2, y t3 3t dy dy Horizontal tangents: 3t2 3 0 when t ±1. Horizontal tangents: 3t2 3 0 when t ±1. dt dt Points: 0, 2, 2, 2 Points: 2, 2, 4, 2 dx dx 1 Vertical tangents:1 0; none Vertical tangents: 2t 1 0 when t . dt dt 2

3 7 11 (2, 2) Point: , 4 8

−4 5

(0,− 2) −3

31. x 3 cos , y 3 sin 32. x cos , y 2 sin 2 dy 3 dy Horizontal tangents: 3 cos 0 when , . Horizontal tangents: 4 cos 2 0 when d 2 2 d 3 5 7 Points: 0, 3, 0, 3 , , , . 4 4 4 4 dx Vertical tangents:3 sin 0 when 0, . 2 2 2 2 d Points: , 2 , , 2 , , 2 , , 2 2 2 2 2 Points: 3, 0, 3, 0 dx 4 Vertical tangents: sin 0 when 0, . (0, 3) d Points: 1, 0, 1, 0 −6 6 (−3, 0) (3, 0)

(0,− 3) −4

33. x 4 2 cos , y 1 sin 34. x 4 cos2 , y 2 sin dy 3 dy 3 Horizontal tangents: cos 0 when , . Horizontal tangents: 2 cos 0 when , . d 2 2 d 2 2 Points: 4, 0, 4, 2 Since dxd 0 at 2 and 32, exclude them. dx dx Vertical tangents:2 sin 0 when 0, . Vertical tangents:8 cos sin 0 when d d Points: 6, 1, 2, 1 0, .

1 Point: 4, 0 (4, 0) 0 8 (2,− 1) (6,− 1)

(4,− 2)

−3

35. x sec , y tan 4 dy 2 Horizontal tangents:sec 0; none −6 6 d (−1, 0) (1, 0) dx Vertical tangents: sec tan 0 when x 0, . − d 4 Points: 1, 0, 1, 0 Section 10.3 Parametric Equations and Calculus 397

36. x cos2 , y cos 37. x t2, y t3 t dy dy 3t2 1 Horizontal tangents:sin 0 when x 0, . d dx 2t Since dxd 0 at these values, exclude them. 2t6t 3t2 12 d2y 4t2 dx Vertical tangents:2 cos sin 0 when 2 2t d dx 6t2 2 1 3t2 3 , . 3 3 2 2 8t 4t Exclude 0, . Concave upward for t > 0 Point: 0, 0 Concave downward for t < 0

38. x 2 t2, y t2 t3 39. x 2t ln t, y 2t ln t, t > 0 dy 2t 3t2 3 dy 2 1t 2t 1 1 t dx 2t 2 dx 2 1t 2t 1 d2y 32 3 d2y 2t 12 2t 12 1 2 dx2 2t 4t dx2 2t 12 t

Concave upward for t > 0 4 t 2t 12 2t 1 Concave downward for t < 0 4t 2t 13 d2y Because t > 0, > 0 dx2

Concave upward for t > 0

40. x t2, y ln t, t > 0 41. x sin t, y cos t, 0 < t < dy 1t 1 dy sin t tan t dx 2t 2t2 dx cos t d2y 1t3 1 d2y sec2t 1 dx2 2t 2t4 dx2 cos t cos3t

d2y Concave upward on 2 < t < Because t > 0, < 0 dx2 Concave downward on 0 < t < 2 Concave downward for t > 0

42. x 2 cos t, y sin t, 0 < t < 2 43. x 2t t2, y 2t32, 1 ≤ t ≤ 2 dy cos t 1 dx dy cot t 2 2t, 3t12 dx 2 sin t 2 dt dt d2y 12 csc2 t 1 b dx 2 dy 2 s dt 2 3 dx 2 sin t 4 sin t a dt dt 2 Concave upward on < t < 2 2 2t2 3t122 dt Concave downward on 0 < t < 1 2 4 8t 4t2 9t dt 1 2 4t2 t 4 dt 1 398 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

44. x ln t, y t 1, 1 ≤ t ≤ 6 45. x et 2, y 2t 1, 2 ≤ t ≤ 2 dx 1 dy dx dy , 1 et, 2 dt t dt dt dt

b dx 2 dy 2 b dx 2 dy 2 s dt s dt a dt dt a dt dt 6 1 2 2t 2 1 dt e 4 dt 1 t 2

46. x t sin t, y t cos t, 0 ≤ t ≤ 47. x t2, y 2t, 0 ≤ t ≤ 2 dx dy dx dy dx 2 dy 2 1 cos t, 1 sin t 2t, 2, 4t2 4 4t2 1 dt dt dt dt dt dt

b dx 2 dy 2 2 s dt s 2 t2 1 dt a dt dt 0 2 2 2 2 2 t t 1 lnt t 1 1 cos t 1 sin t dt 0 0 25 ln2 5 5.916 3 2 cos t 2 sin t dt 0

48. x t2 1, y 4t3 3, 1 ≤ t ≤ 0 49. x et cos t, y et sin t, 0 ≤ t ≤ 2 dx dy dx 2 dy 2 2t, 12t2, 4t2 144t4 dx dy dt dt dt dt etsin t cos t, etcos t sin t dt dt 0 0 2 4 2 2 2 2 s 4t 144t dt 2t 1 36t dt dx dy 1 1 s dt 0 dt dt 2 32 0 1 36t 1 32 2 2 1 37 4.149 54 54 2t t 1 2e dt 2 e 1 dt 0 0

2 2et 21 e2 1.12 0

1 dx 1 dy 50. x arcsin t, y ln1 t 2, 0 ≤ t ≤ 51. x t, y 3t 1, , 3 2 dt 2t dt dx 1 dy 1 2t t 1 1 1 1 1 36t s 9 dt dt 2, 2 2 dt 1 t dt 2 1 t 1 t 0 4t 2 0 t 1 2 dx 2 dy 2 1 6 s dt 1 u2 du 0 dt dt 6 0 1 2 1 1 2 1 1 6 dt dt 2 2 22 2 ln 1 u u u 1 u 0 1 t 0 1 t 12 0 1 t 1 12 1 ln ln 37 6 6 37 3.249 2 t 1 0 12 1 1 1 3 ln ln3 0.549 u 6t, du dt 2 3 2 t Section 10.3 Parametric Equations and Calculus 399

t5 1 dx dy t4 1 dx 52. x t, y , 1, 53. x a cos3 , y a sin3 , 3a cos2 sin , 10 6t3 dt dt 2 2t4 d 2 t 4 1 2 dy S 1 dt 3a sin2 cos 4 1 2 2t d 2 t4 1 2 2 2 4 2 2 4 2 4 dt s 4 9a cos sin 9a sin cos d 1 2 2t 0 2 t4 1 2 2 2 4 dt 12a sin cos cos sin d 1 2 2t 0 5 2 2 2 t 1 779 3 6a sin 2 d 3a cos 2 6a 10 6t 1 240 0 0

dx dy 54. x a cos , y a sin , a sin , a cos 55. x a sin , y a1 cos , d d 2 dx dy a1 cos , a sin S 4 a2 sin2 a2 cos2 d d d 0 2 2 2 2 2 2 s 2 a 1 cos a sin d 4a d 4a 2 a 0 0 0 22a 1 cos d 0 sin 22a d 0 1 cos

42a1 cos 8a 0

dx 56. x cos sin , y sin cos , cos 57. x 90 cos 30t, y 90 sin 30t 16t2 d (a) 35 dy sin d

2 S 2 cos2 2 sin2 d 0 0 240 0 2 2 2 d 2 2 2 0 45 0 (b) Range: 219.2 ft, t 16 dx dy (c) 90 cos 30, 90 sin 3032t dt dt 45 y 0 for t . 16

4516 s 90 cos 302 90 sin 3032t2 dt 0 230.8 ft 400 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

90 58. y 0 ⇒ 90 sin t 16t2 ⇒ t 0, sin 16 90 x 90 cos t 90 cos sin 16 902 902 sin cos sin 2 16 32 902 x 2 cos 2 0 ⇒ 32 4 By the First Derivative Test, 45 maximizes the range x 253.125 feet. 4 dx dy To maximize the arc length, we have 90 cos , 90 sin 32t. dt dt

9016sin s 90 cos 2 90 sin 32t2 dt 0 2025 2025 1 sin sin cos2 ln 8 16 1 sin Using a graphing utility, we see that s is a maximum of approximately 303.67 feet at 0.985556.5.

4t 4t2 59. x , y 1 t3 1 t3 dy 1 t38t 4t23t2 (a) x3 y3 4xy (b) dt 1 t32 4 4t2 t3 0 when t 0 or t 3 2. 1 t32 −6 6 43 2 43 4 Points: 0, 0, , 1.6799, 2.1165 3 3 −4

1 41 2t3 2 4t2 t3 2 1 16 (c) s 2 dt 2 t 8 4t 6 4t 5 4t 3 4t2 1 dt 32 32 34 0 1 t 1 t 0 1 t 1t 8 4t 6 4t 5 4t 3 4t 2 1 8 dt 6.557 32 0 1 t

4 60. x 4 cot , y 4 sin2 , ≤ ≤ tan 2 2 dy (a) 6 (b) 8 sin cos d dx 4 csc2 −6 6 d

− dy 2 0 for 0, ± d 2 (c) Arc length over ≤ t ≤ : 4.5183 4 2 Horizontal tangent at x, y 0, 4 ± 2 (Function is not defined at 0 Section 10.3 Parametric Equations and Calculus 401

61. (a) x t sin t x 2t sin2t (b) The average speed of the particle on the second path is twice the average speed of y 1 cos t y 1 cos2t a particle on the first path. ≤ ≤ ≤ ≤ 0 t 2 0 t 1 1 (c) x 2t sin 2t 3 3 1 y 1 cos 2t The time required for the particle to traverse the same path is t 4. − 3 − 3

−1 −1

62. (a) First particle: x 3 cos t, y 4 sin t,0 ≤ t ≤ 2 (b) There are 4 points of intersection.

4 (c) Suppose at time t that 3 cos t 4 sin t and 4 sin t 3 cos t −6 6 3 3 tan t 4 and tan t 4

−4 Yes, the particles are at the same place at the same time 3 for tan t 4. t 0.6435, 3.7851. The intersection points Second particle: x 4 sin t, y 3 cos t, 0 ≤ t ≤ 2 are 2.4, 2.4 and 2.4, 2.4

4 (d) The curves intersect twice, but not at the same time.

− 66

− 4

dx 1 dx 1 63. x 4t, 4 64. x t2, t dt 4 dt 2 dy dy y t 1, 1 y t 2, 1 dt dt 2 4 t2 S 2 t 142 12 dt S t 2 1 dt 0 0 4 2 4 2 17t 1 dt 817 t 2t2 4 dt 0 0 103.6249 159.6264

dx dx 65. x cos2 , 2 cos sin 66. x sin , 1 cos d d dy dy y cos , sin y cos , 1 sin d d

2 2 S 2 cos 4 cos2 sin2 sin2 d S 2 cos 1 cos 2 1 sin 2 d 0 0 5.3304 2 2 cos 3 2 cos 2 sin d 0 23.2433

dx dy 67. x t, y 2t, 1, 2 dt dt

4 4 4 4 (a) S 2 2t1 4 dt 45 t dt (b) S 2 t1 4 dt 25 t dt 0 0 0 0 4 4 25t 2 325 5 t2 165 0 0 402 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

dx dy 68. x t, y 4 2t, 1, 2 dt dt

2 2 2 (a) S 2 4 2t1 4 dt (b) S 2 t1 4 dt 5t2 45 0 0 0 2 254t t 2 85 0

dx dy 1 69. x 4 cos , y 4 sin , 4 sin , 4 cos 70. x t3, y t 1, 1 ≤ t ≤ 2, y-axis d d 3 2 dx dy S 2 4 cos 4 sin 2 4 cos 2 d t2, 1 0 dt dt 2 2 2 1 2 32 cos d 32 sin 32 S 2 t3t4 1 dt x4 132 0 0 1 3 9 1 1732 232 23.48 9

dx dy 71. x a cos3 , y a sin3 , 3a cos2 sin , 3a sin2 cos d d 2 2 12a2 2 12 S 4 a sin3 9a2 cos4 sin2 9a2 sin4 cos2 d 12a2 sin4 cos d sin5 a2 0 0 5 0 5

dx dy 72. x a cos , y b sin , a sin , b cos d d

2 (a) S 4 b sin a2 sin2 b2 cos2 d 0 2 a2 b2 4ab 2 2 2 2 4 ab sin 1 2 cos d e sin 1 e cos d 0 a e 0 2ab 2 2ab e cos 1 e2 cos2 arcsine cos e1 e2 arcsine e 0 e 2a2b a2 b2 ab 2b2 arcsin 2b2 2 arcsine a2 b2 a e a2 b2 c e : eccentricity a a

2 (b) S 4 a cos a2 sin2 b2 cos2 d 0 2 4a 2 4 a cos b2 c2 sin2 d c cos b2 c2 sin2 d 0 c 0 2a 2 c sin b2 c2 sin2 b2 lnc sin b2 c2 sin2 c 0 2a c b2 c2 b2 lnc b2 c2 b2 ln b c 2ab2 a a2 b2 b2 1 e 2a2 ln 2a2 ln a2 b2 b e 1 e

dy dydt 73. 74. (a) 0 dx dxdt (b) 4 See Theorem 10.7. Section 10.3 Parametric Equations and Calculus 403

b dx 2 dy 2 75. One possible answer is the graph 76. One possible answer is the graph 77. s dt dt dt given by given by a See Theorem 10.8. x t, y t. x t, y t.

y y

4 4 3 2 2 1 x −4 −2 2 4 x −4 −3 −2 1 2 3 4 −2 −2 −3 −4 −4

b dx 2 dy 2 78. (a) S 2 gt dt 79. Let y be a continuous function of x on a ≤ x ≤ b. a dt dt Suppose that x f t , y g t , and f t1 a, f t2 b. b dx 2 dy 2 Then using integration by substitution,dx ft dt and (b) S 2 f t dt a dt dt b t2 y dx gtft dt.

a t1

80. x r cos , y r sin S 2 r sin r2 sin2 r2 cos2 d y 0 2 2 r sin d θ 0 2r2 cos x 0 2r21 cos

81. x 2 sin2 y y 2 sin2 tan 2

π 1 dx ≤ θ < 4 sin cos 0 2 d x −2 −1 1 2 2 − A 2 sin2 tan 4 sin cos d 8 sin4 d 1 0 0 −2 sin3 cos 3 3 2 3 8 sin cos 4 8 8 0 2

dx 82. x 2 cot , y 2 sin2 , 2 csc2 d

0 0 0 A 2 2 sin2 2 csc2 d 8 d 8 4 2 2 2

3 2 2 83. ab is area of ellipse (d). 84. 8 a is area of asteroid (b). 85. 6 a is area of cardioid (f).

2 8 86. 2a is area of deltoid (c). 87. 3ab is area of hourglass (a). 88. 2ab is area of teardrop (e). 404 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

89. x t, y 4 t, 0 < t < 4 2 4 1 1 4 1 2 4 16 A y dx 4 t dt 4t12 t12 dt 8t tt 0 0 2t 2 0 2 3 0 3 1 2 3 4 1 3 4 3 t2 4 3 x yx dx 4 tt dt 4 t dt 4t A 0 16 0 2t 32 0 32 2 0 4 1 2 y2 3 4 1 3 4 3 16 2 4 8 y dx 4 t2 dt 16t128t12 t32 dt 32t tt t2t A 0 2 32 0 2t 64 0 64 3 5 0 5 3 8 x, y , 4 5

dx 1 90. x 4 t, y t, , 0 ≤ t ≤ 4 dt 24 t 0 1 2 1 u 2 A t dt 4 u2 du u4 u2 4 arcsin 4 24 t 0 2 2 0 Let u 4 t, then du 124 t dt and t 4 u2. 1 0 1 1 0 1 2 0 8 32 x 4 t t dt t dt t 4 24 t 2 4 2 3 4 3 0 0 0 1 2 1 1 t 1 2 8 t 8 y t dt dt 4 t 2 4 24 t 4 4 4 t 4 3 4 3 8 8 x, y , 3 3

dx dx 91. x 3 cos , y 3 sin , 3 sin 92. x cos , y 3 sin , sin d d

0 0 V 2 3 sin 23 sin d V 2 3 sin 2sin d 2 2 0 0 54 sin3 d 18 sin3 d 2 2 0 cos3 0 54 1 cos2 sin d 18cos 12 2 3 2 cos3 0 54cos 36 (Sphere) 3 2

93. x a sin , y a1 cos dy dx (a) a sin , a1 cos d d dy a sin sin dx a1 cos 1 cos d2y 1 cos cos sin sin a1 cos dx2 1 cos 2 cos 1 1 a1 cos 3 acos 12 1 3 dy 12 (b) At , x a , y a1 , 2 3. 6 6 2 2 dx 1 32 3 1 Tangent line: y a1 2 3x a 2 6 2 —CONTINUED— Section 10.3 Parametric Equations and Calculus 405

93. —CONTINUED— dy sin 2 ⇒ 2 2 2 2 (c) 0 sin 0, 1 cos 0 (e) s a sin a 1 cos d dx 1 cos 0 Points of horizontal tangency: x, y a2n 1, 2a 2 a 2 2 cos d (d) Concave downward on all open -intervals: 0 . . ., 2, 0, 0, 2, 2, 4, . . . 2 a 4 sin22 d 0 2 2a sin d 0 2

2 4a cos 8a 2 0

2 1 3 94. x t 3, y 3t 3t 3 2 2 (a) 6 (d) s 12t 3 t dt 3 3 0 18 t4 6t2 9 12t2 dt 3 3 −6 t2 32 dt 3 dx dy dy 3 t2 (b) 23t, 3 t2, 3 dt dt dx 23t t3 3 dt 36 3 d2y 23t2t 3 t223 23t 3 dx2 12t2 1 3 3 (e) S 2 3t 3t t 3 dt 81 0 23t2 63 t2 3 12t223t 12t3 dy 2 3 (c)x, y 3, 8 at t 1. 3 dx 23 3 8 3 y x 3 3 3 3 5 y x 3 3

95. x t u r cos r sin y rcos sin y v w r sin r cos r sin cos θ r θ r w v (,x y ) θ u x t 406 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

96. Let’s focus on the region above the x -axis. From Exercise 95, y the equation of the involute from 1, 0 to 1, is (−1, π (

π x cos sin ( ,1 ( A B 2 y sin cos DC x (−1 − π , 0) (−1, 0) (1, 0) 0 ≤ ≤ . At 1, , the string is fully extended and has length . So, the area 12 13 of region A is 4 4 . We now need to find the area of region B. dx sin sin cos cos 0 ⇒ . 0 is cusp. d 2 Hence, the far right point on the involute is 2, 1. The area of the region B C D is given by

2 2 0 y dx y dx y dx 0 where y sin cos and dx cos d. Thus, we can calculate 0 sin cos cos d 2 3. 6 Since the area of C D is 2, we have 1 5 Total area covered 2 3 2 3 3. 4 6 2 6

97. (a) 2

−3 3

−2

1 t2 2t (b) x , y , 20 ≤ t ≤ 20 1 t2 1 t2

The graph for < t < is the circle x2 y2 1, except the point 1, 0. 1 t2 2 2t 2 1 2t2 t4 4t2 1 t22 Verify: x2 y2 1 1 t2 1 t2 1 t22 1 t22 (c) As t increases from 20 to 0, the speed increases, and as t increases from 0 to 20, the speed decreases.

12 144 x2 t t 98. (a) y 12 ln 144 x2 (b) x 12 sech , y t 12 tanh , 0 ≤ t x 12 12

0 < x ≤ 12 60

0 12 0

0 12 0 Same as the graph in (a), but has the advantage of showing the position of the object and any given time t. —CONTINUED— Section 10.4 Polar Coordinates and Polar Graphs 407

98. —CONTINUED— dy 1 sech2t12 t y (c) sinh dx secht12 tant12 12 24 (0,y0 )

t0 t0 t0 Tangent line: y t 12 tanh sinh x 12 sech 16 0 12 12 12 12 (,x y ) t0 8 y t0 sinh x 12 4 x y-intercept: 0, t0 246 10 12 t 2 t 2 Distance between 0, t and x, y: d 12 sech 0 12 tanh 0 12 0 12 12 d 12 for any t ≥ 0. d gt d 2y dt ft ftgt gtft 99. False. 100. False. Both dxdt and dydt are zero when t 0. By dx2 ft ft3 eliminating the parameter, we have y x23 which does not have a horizontal tangent at the origin.

Section 10.4 Polar Coordinates and Polar Graphs

7 1. 4, 2. 2, 3. 4, 2 4 3 7 x 4 cos 0 x 2 cos 2 x 4 cos 2 2 4 3 7 y 4 sin 4 y 2 sin 2 y 4 sin 23 2 4 3 x, y 0, 4 x, y 2, 2 x, y 2, 23

π π π 2 2 2 − π ()2, 2 4, 3 ()6 π −−4, ()3 0 12

0 123 0 1

7 4. 0, 5. 2, 2.36 6. 3, 1.57 6 x 2 cos2.36 1.004 x 3 cos1.57 0.0024 7 x 0 cos 0 6 y 2 sin2.36 0.996 y 3 sin1.57 3 7 x, y 1.004, 0.996 x, y 0.0024, 3 y 0 sin 0 6 π π 2 2 x, y 0, 0

π ()2, 2.36 (−0.0024, 3) 2

0 1 (0, 0) 0 0 1 12 408 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

3 11 7. r, 5, 8. r, 2, 9. r, 3.5, 2.5 4 6 x, y 2.804, 2.095 x, y 3.5355, 3.5355 x, y 1.7321, 1 y y y 1 4 2 − (−3.54, 3.54) ( 1.7321, 1) x 3 −1 132 1 2 −1 x 1 −2 −1 1 2 −2 (2.804,− 2.095) x −4 −3 −2 −1 1 −1 −3 −1 −2

10. r, 8.25, 1.3 11. x, y 1, 1 12. x, y 0, 5 x, y 2.2069, 7.9494 r ±2 r ±5

y tan 1 tan undefined 8 (2.2069, 7.9494) 5 5 3 3 6 , , 2, , 2, , , 5, , 5, 4 4 4 4 4 2 2 2 2 2 y y x −2 −1 12 − x 2 −2 −1 123 −4 2 −1 −6 −2 −8 (1, 1) 1 −3

−4

x −5 (0,− 5) 12

13. x, y 3, 4 14. x, y 4, 2 r ±9 16 ±5 r ±16 4 ±25 tan 4 2 1 3 tan 4 2 2.214, 5.356, 5, 2.214, 5, 5.356 y 0.464 5 2 5, 0.464 , 2 5, 2.678

(−3, 4) 4 y

3 x 12345 2 −1 1 −2 x (4, −2) −4 −3 −2 −11 −3

−4

−5

15. x, y 3, 1 y 16 x, y 3, 3 y

2 r 3 1 2 1 r 9 3 23 1 x tan 3 3 tan 3 3 x −112 −11234 −1 −1 ( 3 , −1) −2 ) 3, − 3 ) r, 2, 11 6 −2 r, 2 3, 11 6 −3 2, 56 23, 56 Section 10.4 Polar Coordinates and Polar Graphs 409

17. x, y 3, 2 18. x, y 32, 32 r, 3.606, 0.588 r, 6, 0.785

5 4 19. x, y 2, 3 20. x, y 0, 5 r, 2.833, 0.490 r, 5, 1.571

21. (a) x, y 4, 3.5 y (b) r, 4, 3.5 π 2 4 (4, 3.5)

3 0 1

2 (4, 3.5) 1

x 1234

22. (a) Moving horizontally, the x -coordinate changes. Moving vertically, the y -coordinate changes. (b) Both r and values change. (c) In polar mode, horizontal (or vertical) changes result in changes in both r and .

23. r 2 sin circle 24. r 4 cos 2 25. r 31 cos 26. r 2 sec Matches (c) Rose curve Cardioid Line Matches (b) Matches (a) Matches (d)

27. x2 y2 a2 π 28. x2 y2 2ax 0 π 2 2 r a r2 2ar cos 0 rr 2a cos 0 0 0 a r 2a cos a 2a

29. y 4 π 30. x 10 π 2 2 r sin 4 r cos 10 r 4 csc r 10 sec 0 2468 12 0 2 4

π 31. 3 x y 2 0 2 3r cos r sin 2 0 r3 cos sin 2 2 r 0 3 cos sin 1 2 410 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

32. xy 4 33. y2 9x r cos r sin 4 r2 sin2 9r cos r2 4 sec csc 9 cos r sin2 8 csc 2 2 π r 9 csc cos 2 π 2

0 24 0 13572 46

34. x2 y22 9x2 y2 0 35. r 3 r22 9r2 cos2 r2 sin2 0 r2 9 r2r2 9cos 2 0 x2 y2 9 r2 9 cos 2 y

π 2 2

1 x 2 1 1 2 1

0 2 12

36. r 2 37. r sin 38. r 5 cos r2 4 r2 r sin r2 5r cos x2 y2 4 x2 y2 y x2 y2 5x y 1 2 1 25 25 x2 y x2 5x y2 2 4 4 4

2 2 2 2 1 x y y 0 5 5 x y2 2 2 x y −1 1 y −1 4 3

1 2 2 1 x −2 −112346 x 1 1 − 2 2 2 −3 −4

39. r y 2π tan r tan π

2 2 y tan x y x x π 2π − π y x2 y2 arctan x −2π Section 10.4 Polar Coordinates and Polar Graphs 411

5 40. 41. r 3 sec 42. r 2 csc 6 r cos 3 r sin 2 5 tan tan 6 x 3 y 2 y 3 x 3 0 y 2 0 x 3 y y 3 y x 3 3 3

y 2

2 1 1

1 x x 1 2 −1 1 2 x −2 −1 1 2

−1

−2

43. r 3 4 cos 44. r 51 2 sin 45. r 2 sin 0 ≤ < 2 0 ≤ < 2 0 ≤ < 2

6 3 4

−10 10

−12 6 −4 5

−6 −18 −2

2 2 46. r 4 3 cos 47. r 48. r 1 cos 4 3 sin 0 ≤ < 2 Traced out once on < < Traced out once on 0 ≤ ≤ 2 6 5 3

−4 10 −10 5

−3 3

−6 − 5 −1

3 5 49. r 2 cos 50. r 3 sin 51. r2 4 sin 2 2 2 r 2sin 2 0 ≤ < 4 0 ≤ < 4 1 r 2sin 2 2 4 2 0 ≤ < −3 3 −6 6 2

−2 −4

−33

−2 412 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

1 2 52. r .

1 1 Graph as r , r . 1 2 It is traced out once on 0, .

1.5

−2 2

−1.5

53. r 2h cos k sin r2 2rh cos k sin r2 2hr cos kr sin x2 y2 2hx ky x2 y2 2hx 2ky 0 x2 2hx h2 y2 2ky k2 0 h2 k2 Radius: h2 k2 x h2 y k2 h2 k2 Center: h, k

54. (a) The rectangular coordinates of r1, 1 are r1 cos 1, r1 sin 1 . The rectangular coordinates of r2, 2 are r2 cos 2, r2 sin 2 . 2 2 2 d x2 x1 y2 y1 2 2 r2 cos 2 r1 cos 1 r2 sin 2 r1 sin 1 2 2 2 2 2 2 2 2 2 r2 cos 2 2r1r2 cos 1 cos 2 r1 cos 1 r2 sin 2 2r1r2 sin 1 sin 2 r1 sin 1 2 2 2 2 2 2 r2 cos 2 sin 2 r1 cos 1 sin 1 2 r1r2 cos 1 cos 2 sin 1 sin 2 2 2 r1 r2 2r1r2 cos 1 2 2 2 d r1 r2 2r1r2 cos 1 2 (b) If 1 2, the points lie on the same line passing through the origin. In this case, 2 2 d r1 r2 2r1r2 cos 0 2 r1 r2 r1 r2. 2 2 (c) If 1 2 90 , then cos 1 2 0 and d r1 r2 , the Pythagorean Theorem! (d) Many answers are possible. For example, consider the two points r1, 1 1, 0 and r2, 2 2, 2 . d 1 22 212 cos0 5 2 5 Using r , 1, and r , 2, 52 , d 12 22 212 cos 5. 1 1 2 2 2 You always obtain the same distance. Section 10.4 Polar Coordinates and Polar Graphs 413

2 7 55. 4, , 2, 56. 10, , 3, 3 6 6 2 7 d 42 22 242 cos d 102 32 2103 cos 3 6 6 20 16 cos 25 4.5 109 60 cos 109 303 7.6 2 6

57. 2, 0.5, 7, 1.2 58. 4, 2.5, 12, 1 d 22 72 227 cos0.5 1.2 d 42 122 2412 cos2.5 1 53 28 cos0.7 5.6 160 96 cos 1.5 12.3

59. r 2 3 sin 60. r 21 sin dy 3 cos sin cos 2 3 sin dy 2 cos sin 2 cos 1 sin dx 3 cos cos sin 2 3 sin dx 2 cos cos 2 sin 1 sin 2 cos 3 sin 1 2 cos 3 sin 1 dy At 2, 0, 1. 3 cos 2 2 sin 6 cos2 2 sin 3 dx dy 7 dy At 5, , 0. At 3, , is undefined. 2 dx 6 dx dy 2 3 dy At 2, , . At 4, , 0. dx 3 2 dx 3 dy At 1, , 0. 2 dx

61. (a), (b) r 31 cos 62. (a), (b) r 3 2 cos

4 4

− 8 4 −8 4

− 4 −4 r, 3, ⇒ x, y 0, 3 r, 1, 0 ⇒ x, y 1, 0 2 Tangent line: x 1 Tangent line: y 3 1x 0 dy y x 3 (c) At 0, does not exist (vertical tangent). dx dy (c) At , 1.0. 2 dx

5 dy 63. (a), (b) r 3 sin (c) At , 3 1.732. 3 dx

−4 5

−1

33 33 9 r, , ⇒ x, y , 2 3 4 4 9 33 Tangent line: y 3 x 4 4 9 y 3x 2 414 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

64. (a), (b) r 4 65. r 1 sin dy 6 1 sin cos cos sin d −8 8 cos 1 2 sin 0 1 3 5 cos 0 or sin ⇒ , , , −6 2 2 2 6 6 3 1 1 5 Horizontal tangents: 2, , , , , r, 4, ⇒ x, y 22, 22 2 2 6 2 6 4 dx 1 sin sin cos cos Tangent line: y 2 2 1x 2 2 d y x 4 2 sin sin2 sin2 1 dy (c) At , 1. 2 sin2 sin 1 4 dx 2 sin 1sin 1 0 1 7 11 sin 1 or sin ⇒ , , 2 2 6 6 3 7 3 11 Vertical tangents: , , , 2 6 2 6

66. r a sin 67. r 2 csc 3 dy dy a sin cos a cos sin 2 csc 3 cos 2 csc cot sin d d 2a sin cos 0 3 cos 0 3 3 0, , , , 2 2 2 2 dx 3 a sin2 a cos2 a1 2 sin2 0 Horizontal: 5, , 1, d 2 2 1 3 5 7 sin ± , , , , 2 4 4 4 4 Horizontal: 0, 0, a, 2 a2 a2 3 Vertical: , , , 2 4 2 4

68. r a sin cos2 69. r 4 sin cos2

dy 2 a sin cos3 2a sin2 cos a cos3 sin d

2asin cos3 sin3 cos −3 3 2a sin cos cos2 sin2 0 −2 3 0, tan2 1, , 4 4 Horizontal tangents: 2a 2a 3 r, 0, 0, 1.4142, 0.7854, 1.4142, 2.3562 Horizontal: , , , , 0, 0 4 4 4 4 Section 10.4 Polar Coordinates and Polar Graphs 415

70. r 3 cos 2 sec 71. r 2 csc 5

2 10

−2 4 −12 12

−2 −6

Horizontal tangents: r, 2.061, ±0.452 3 Horizontal tangents: r, 7, , 3, 2 2

72. r 2 cos3 2 73. r 3 sin π 2

2 r2 3r sin x2 y 2 3y −3 3 0 3 2 9 123 x 2 y 2 4 −2 3 Circle r Horizontal tangents: 2 r, 1.894, 0.776, 1.755, 2.594, 1.998, 1.442, 3 Center: 0, 0.423, 0.072 2 Tangent at the pole: 0

74. r 3 cos π 75. r 21 sin π 2 2 r2 3r cos Cardioid 0 1 2 3 x2 y2 3x Symmetric to y -axis, 0 2 3 2 9 12 4 x y2 2 4 3 Circle: r 2 3 Center: , 0 2 Tangent at pole: 2

π 76. r 3 1 cos 2 Cardioid Symmetric to polar axis since r is a function of cos . 0 1 2 0 3 2 3 3 9 r 03 6 2 2 416 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

77. r 2 cos3 Rose curve with three petals π 2 Symmetric to the polar axis 2 Relative extrema: 2, 0, 2, , 2, 3 3 0 2 2 5 0 6 4 3 2 3 6 r 202 2 0 2 0 2

5 Tangents at the pole: , , 6 2 6

78. r sin5 79. r 3 sin 2 Rose curve with five petals Rose curve with four petals Symmetric to Symmetric to the polar axis, , and pole 2 2 Relative extrema occur when 5 Relative extrema: ±3, , ±3, 4 4 dr 3 5 7 9 5 cos5 0 at , , , , . d 10 10 10 10 10 Tangents at the pole: 0, 2 2 3 4 Tangents at the pole: 0, , , , 5 5 5 5 , 32 give the same tangents.

π π 2 2

0 0 1 3

80. r 3 cos 2 π 2 Rose curve with four petals Symmetric to the polar axis, , and pole 0 2 2 3 Relative extrema: 3, 0, 3, , 3, , 3, 2 2 3 Tangents at the pole: , 4 4 5 7 and given the same tangents. 4 4

81. r 5 π 82. r 2 π 2 2 Circle radius: 5 Circle radius: 2 x2 y2 25 x2 y2 4 0 0 2 4 6 1 Section 10.4 Polar Coordinates and Polar Graphs 417

83. r 41 cos π 84. r 1 sin π 2 2 Cardioid Cardioid

0 2104 6 0 12

85. r 3 2 cos π 86. r 5 4 sin π 2 2 Limaçon Limaçon 0 24 Symmetric to polar axis Symmetric to 0 2 2 2 0 3 2 3 0 2 6 6 2 r 12 3 4 5 r 97531

6 87. r 3 csc 88. r 2 sin 3 cos r sin 3 2 r sin 3r cos 6 y 3 2 y 3x 6 Horizontal line Line π 2 π 2

0 1 2 0 1

1 89. r 2 90. r Spiral of Archimedes Hyperbolic spiral Symmetric to 2 3 5 3 4 2 4 4 2 3 5 3 4 2 4 1 4 2 0 r 4 2 4 4 2 3 5 3 3 5 r 03 2 π 2 2 2 2

Tangent at the pole: 0

π 2

0 1

0 132 418 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

91. r2 4 cos2 π 2 r 2cos 2, 0 ≤ ≤ 2 Lemniscate 0 Symmetric to the polar axis, , and pole 1 2 Relative extrema: ±2, 0

0 6 4 r ±2 ±2 0

3 Tangents at the pole: , 4 4

2 π 92. r 4 sin 2 Lemniscate Symmetric to the polar axis, , and pole 2 0 2 Relative extrema: ±2, 2

5 0 6 2 6 r 00±2 ±2 ±2

Tangent at the pole: 0

93. Since 94. Since 1 1 r 2 sec 2 , r 2 csc 2 , cos sin the graph has polar axis symmetry and the tangents the graphs has symmetry with respect to 2. at the pole are Furthermore, r ⇒ as ⇒ 0 , . − 3 3 x =1 4 r ⇒ as ⇒ . Furthermore, 1 r r − ⇒ ⇒ 6 6 Also, r 2 2 2 r as sin sin y 2 ry 2y r − r ⇒ as ⇒ . 4 2 2y r . y 1 1 r r Also, r 2 2 2 cos r cos x Thus,r ⇒ ± as y ⇒ 1.

rx 2x r 4 2x r . 1 x −4 4 Thus,r ⇒ ± as x ⇒ 1. y = 1 −2 Section 10.4 Polar Coordinates and Polar Graphs 419

2 95. r 96. r 2 cos 2 sec Strophoid Hyperbolic spiral r ⇒ as ⇒ 0 r ⇒ as ⇒ 2 2 2 2 sin 2 sin r ⇒ r r sin y r ⇒ as ⇒ 2 2 sin 2 y r 2 cos 2 sec 2 2 cos 1 sec r cos 4 cos2 2 2 sin 2 cos lim lim 2 →0 →0 1 x 4 cos2 2

3 lim 4 cos2 2 2 →± y =2 2

x =2− 2

−33

−1 −3 3

−2

97. The rectangular coordinate system consists of all points of the form x, y where x is the directed distance from the y -axis to the point, and y is the directed distance from the x -axis to the point. Every point has a unique representation. The polar coordinate system uses r, to designate the location of a point. r is the directed distance to the origin and is the angle the point makes with the positive x -axis, measured clockwise. Points do not have a unique polar representation.

98. x r cos , y r sin 99. r a, circle y b, line x2 y2 r2, tan x

100. Slope of tangent line to graph of r f at r, is dy f cos fsin . dx f sin fcos If f 0 and f 0, then is tangent at the pole.

101. r 4 sin (a) 0 ≤ ≤ (b) ≤ ≤ (c) ≤ ≤ 2 2 2 2

π π π 2 2 2

0 0 0 12 12 12 420 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

102. r 61 cos (a) 0, r 61 cos (c) 2 9 r 61 cos 2 −9 15 61 cos cos sin sin 2 2 −9 61 sin (b) , r 61 cos 15 4 4

−12 12

− −9 15 3

The graph of r 61 cos is rotated through the −6 angle 2. The graph of r 61 cos is rotated through the angle 4.

π 103. Let the curve r f be rotated by to form the curve r g . 2 If r1, 1 is a point on r f , then r1, 1 is on r g . That is, (,r θ +φ ) g 1 r1 f 1 . Letting , or , we see that 1 1 (,r θ ) g g 1 f 1 f . φ

θ 0

104. (a) sin sin cos cos sin (b) sin sin cos cos sin 2 2 2 sin cos r f sin r f sin 2 f sin f cos 3 3 3 (c) sin sin cos cos sin 2 2 2 cos 3 r f sin f cos 2

105. r 2 sin 2 (a) r 2 sin 2 sin cos (b) r 2 sin 2 cos 2 cos 4 2 2

4 4

−66 −66

−4 −4

—CONTINUED— Section 10.4 Polar Coordinates and Polar Graphs 421

105. —CONTINUED— 3 (c) r 2 sin 2 sin 2 sin (d) r 2 sin 2 cos 2 4 4

−66 −66

−4 −4

106. r 2 sin 2 4 sin cos (a) r 4 sin cos (b) r 4 sin cos 4 sin cos 6 6 2 2

2 2

−3 3 −3 3

−2 −2

2 2 (c) r 4 sin cos (d) r 4 sin cos 4 sin cos 3 3 2 2

−3 3 −3 3

−2 −2

107. (a) r 1 sin π (b) r 1 sin π 2 4 2 Rotate the graph of r 1 sin 0 0 1 2 1 2 through the angle 4.

π 108. By Theorem 9.11, the slope of the tangent line through A and P is 2 Radial line f cos f sin . Polar curve ψ f sin f cos rf=()θ This is equal to Pr= ( ,θ ) tan tan sin cos tan Tangent tan . line 1 tan tan cos sin tan θ 0 A Equating the expressions and cross-multiplying, you obtain f cos f sin cos sin tan sin cos tan f sin f cos f cos2 f cos sin tan f sin cos f sin2 tan f sin2 f sin cos tan f sin cos f cos2 tan f cos2 sin2 f tan cos2 sin2 f r tan . f drd 422 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

r 21 cos r 31 cos 109. tan 110. tan drd 2 sin drd 3 sin 3 1 22 2 2 At , tan is undefined ⇒ . At , tan . 2 4 2 2 3 2 2 arctan 1.178 67.5 2 −6 3 5

−3 −8 2

−5

r 4 sin 2 111. r 2 cos 3 112. tan drd 8 cos 2 r 2 cos 3 1 tan cot 3 sin3 3 drd 6 sin 3 3 At , tan . 6 2 cos3 2 1 3 1 At , tan cot . 3 4 3 4 3 arctan 0.7137 40.89 2 1 arctan 18.4 4 3

2 −6 6 θ −3 3 −4 ψ −2

16 6 1 dr 6 sin 113. r 61 cos ⇒ ψ 1 cos d 1 cos 2 θ 6 −20 22 r 1 cos 1 cos tan dr 6 sin sin −12 d 1 cos 2 1 1 2 2 At , tan 3. 3 3 2 , 60 3

r 5 114. tan undefined ⇒ drd 0 2

−9 9

−6

115. True 116. True 117. True 118. True Section 10.5 Area and Arc Length in Polar Coordinates 423

Section 10.5 Area and Arc Length in Polar Coordinates

1 1 1. A f2 d 2. A f 2 d 2 2 54 1 1 2 sin 2 d cos 22 d 2 2 2 34 2 sin2 d 2

1 1 3. A f 2 d 4. A f 2 d 2 2 1 3 2 1 1 sin 2 d 1 cos 22 d 2 2 2 0

1 2 5. (a) r 8 sin (b) A 2 8 sin 2 d 2 0 π 2 2 64 sin2 d 0 2 32 1 cos 2 d 0 sin 2 2 0 32 16 24 2 0

A 42 16

1 2 6. (a) r 3 cos (b) A 2 3 cos 2 d 2 0 π 2 2 9 cos2 d 0 9 2 1 cos 2 d 2 0 0 24 9 sin 2 2 9 2 2 0 4

3 2 9 A 2 4

6 6 4 4 1 1 1 7. A 2 2 cos 32 d 2 sin 6 8. A 2 6 sin 22 d 36 sin2 2 d 2 6 0 3 0 2 0 0 41 cos 4 36 d 0 2 sin 4 4 18 4 0 9 18 4 2 424 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

1 4 1 10 9. A 2 cos 22 d 10. A 2 cos 52 d 2 0 2 0 1 1 4 1 1 10 sin 4 sin10 2 4 0 8 2 10 0 20

1 2 1 2 11. A 2 1 sin 2 d 12. A 2 1 sin 2 d 2 2 2 0 3 1 2 3 3 1 2 3 8 2 cos sin 2 2 cos sin 2 2 4 2 2 2 4 0 4

1 1 2 13. A 2 1 2 cos 2 d 14. A 2 4 6 sin 2 d 2 23 2 arcsin23 2 33 2 3 4 sin sin 2 16 48 sin 36 sin2 d 23 2 arcsin23 2 2 1 cos 2 16 48 sin 36 d arcsin23 2

−14 2 34 48 cos 9 sin 2 1.7635 arcsin23 −2 2

−8 8

−12

15. The area inside the outer loop is 2 1 2 3 23 4 33 2 1 2 cos 2 d 3 4 sin sin 2 . −14 2 0 0 2 From the result of Exercise 13, the area between the loops is −2 4 33 2 33 A 33. 2 2

16. Four times the area in Exercise 15,A 4 33 . More specifically, we see that the area inside the outer loop is 1 2 2 2 21 2 sin 2 d 4 16 sin 16 sin2 d 8 63. 6 2 6 6 The area inside the inner loop is 1 3 2 2 21 2 sin 2 d 4 63. −4 4 2 76 −1 Thus, the area between the loops is 8 63 4 63 4 123. Section 10.5 Area and Arc Length in Polar Coordinates 425

17. r 1 cos 18. r 31 sin r 1 cos r 31 sin Solving simultaneously, Solving simultaneously, 1 cos 1 cos 3 1 sin 31 sin 2 cos 0 2 sin 0 3 0, . , . 2 2 Replacing r by r and by in the first equation Replacing r by r and by in the first equation and solving, 31 sin 31 sin , sin 1, and solving, 1 cos 1 cos , cos 1, 0. 2. Both curves pass through the pole, 0, 32, Both curves pass through the pole,0, , and 0, 0, and 0, 2, respectively. respectively. Points of intersection: 3, 0, 3, , 0, 0 3 Points of intersection: 1, , 1, , 0, 0 2 2

19. r 1 cos 20. r 2 3 cos r 1 sin r cos Solving simultaneously, Solving simultaneously, 1 cos 1 sin 2 3 cos cos cos sin 1 cos 2 tan 1 5 3 7 , . , . 3 3 4 4 Both curves pass through the pole, (0,arccos 23 ), and Replacing r by r and by in the first equation 0, 2, respectively. and solving, 1 cos 1 sin , sin cos 2, which has no solution. Both curves pass through the pole, 1 1 5 Points of intersection: , , , , 0, 0 0, , and 0, 2, respectively. 2 3 2 3 2 2 3 2 2 7 Points of intersection: , , , , 0, 0 2 4 2 4

21. r 4 5 sin 22. r 1 cos r 3 sin r 3 cos Solving simultaneously, Solving simultaneously, 4 5 sin 3 sin 1 cos 3 cos 1 1 sin cos 2 2 5 5 , . , . 6 6 3 3 Both curves pass through the pole,0, arcsin 45, and 0, 0, Both curves pass through the pole,0, , and 0, 2, respectively. respectively. 3 3 5 3 3 5 Points of intersection: , , , , 0, 0 Points of intersection: , , , , 0, 0 2 6 2 6 2 3 2 3 426 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

23. r π 24. π 2 2 4 2 r 2 r 2 Solving simultaneously, we Line of slope 1 passing 0 0 have 1 through the pole and a circle 13 of radius 2 centered at the 2 2, 4. pole. Points of intersection: Points of intersection: 2, 4, 2, 4 2, , 2, 4 4

25. r 4 sin 2 π r 2 2 r 4 sin 2 is the equation of a rose curve with four petals and is symmetric to the polar axis, 2, and the pole. Also,r 2 is the equation of a circle of radius 2 centered at the pole. 0 Solving simultaneously, 13 4 sin 2 2 5 2 , 6 6 5 , . 12 12 Therefore, the points of intersection for one petal are 2, 12 and 2, 512. By symmetry, the other points of intersection are 2, 712, 2, 1112, 2, 1312, 2, 1712, 2, 1912, and 2, 2312.

π 26. r 3 sin 2 Points of intersection: r 2 csc 17 3 17 3 , arcsin , 2 2 17 3 17 3 0 , arcsin , 12 2 2 3.56, 0.596, 3.56, 2.545

The graph of r 3 sin is a limaçon symmetric to 2, and the graph of r 2 csc is the horizontal line y 2. Therefore, there are two points of intersection. Solving simultaneously, 3 sin 2 csc sin 2 3 sin 2 0 3 ± 17 sin 2 17 3 arcsin 0.596. 2 Section 10.5 Area and Arc Length in Polar Coordinates 427

27. r 2 3 cos θ r = sec 4 2 sec r 2 −4 8 The graph of r 2 3 cos is a limaçon with an inner loop b > a and is symmetric to the polar axis. The graph of r sec 2 is the vertical line x 12. Therefore, there −4 are four points of intersection. Solving simultaneously, r =2+3 cosθ sec 2 3 cos 2 6 cos2 4 cos 1 0 2 ± 10 cos 6 2 10 arccos 1.376 6 2 10 arccos 2.6068. 6 Points of intersection: 0.581, ±2.607, 2.581, ±1.376

28. r 31 cos r = 6 1 − cos θ 6 5 r 1 cos

− The graph of r 31 cos is a cardioid with polar axis symmetry. The graph of 10 5 r 61 cos −5 r = 3(1− cos θ ) is a parabola with focus at the pole, vertex3, , and polar axis symmetry. Therefore, there are two points of intersection. Solving simultaneously, 6 3 1 cos 1 cos 1 cos 2 2 cos 1 ± 2 arccos 1 2 . Points of intersection: 32, arccos 1 2 4.243, 1.998, 32, 2 arccos1 2 4.243, 4.285

29. r cos 30. r 4 sin r 2 3 sin r 21 sin Points of intersection: Points of intersection: 0, 0, 4, 2 0, 0, 0.935, 0.363, 0.535, 1.006 The graphs reach the pole at different times ( values). The graphs reach the pole at different times ( values). 6 r = cos θ 1 r = 4 sin θ

−4 5

−6 6

−2 r = 2 (1 + sin θ ) −5 r = 2− 3 sin θ 428 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

31. From Exercise 25, the points of intersection for one petal are 2, 12 and 2, 512. The area within one petal is 1 12 1 5 12 1 2 2 2 2 4 sin 2 d 2 d 4 sin 2 d θ r = 2 4 r = 4 sin 2 2 0 2 12 2 512 12 512 2 16 sin 2 d 2 d (by symmetry of the petal) −6 6 0 12

12 512 1 4 8 sin 4 2 3. −4 4 0 12 3 4 16 4 Total area 4 3 43 4 33 3 3 3

1 2 1 2 32. A 4 91 sin 2 d 33. A 4 3 2 sin 2 d 2 0 2 0 2 9 2 18 1 sin 2 d 3 8 211 12 cos sin2 11 24 0 2 0 − θ (from Exercise 14) 6 r = 3 + 2 sin

−9 9

−7 7

− 6 r = 3 − 2 sin θ

−7

1 6 1 2 34. r 5 3 sin and r 5 3 cos intersect at 35. A 2 4 sin 2 d 22 d 2 2 4 and 54. 0 6 6 2 54 1 1 1 2 16 sin 2 4 A 2 5 3 sin d 2 4 0 6 2 4 8 2 59 9 5 4 23 4 33 30 cos sin 2 3 3 2 4 4 r = 4 sin θ 59 5 2 9 59 2 9 5 30 30 2 4 2 4 2 4 2 4 59 302 50.251 −66 2

−3 8 r = 2

−12 12

−8

1 2 1 2 1 a2 36. A 2 3 sin 2 d 2 sin 2 d 37. A 2 a1 cos 2 d 2 6 2 6 2 0 4 2 3 sin 2 a2 4 cos 2 4 sin d a2 2 sin 6 2 4 0 4 2 3a2 a2 5a2 2 sin 2 4 cos 3 3 6 2 4 4 4

−4 4

−4 Section 10.5 Area and Arc Length in Polar Coordinates 429

a2 1 38. Area Area of r 2a cos Area of sector 39. A a1 cos 2 d twice area between r 2a cos and the lines 8 2 2 2 2 a a 3 cos 2 , . 2 cos d 3 2 8 2 2 2 2 2 2 1 2 a a 3 sin 2 2 2 2 2 sin A a a 2 2a cos d 8 2 2 4 2 3 2 3 2 2 2 2 2 a a 3 3 a 2 a 2 2 2 2a 1 cos 2 d 8 2 2 4 2 3 3 π 2 2a sin 2 2 2 2a2 3 2 3 2a2 3 2a2 33a2 2a2 3 2 3 4 6 0 a 2a

π π θ 2 = 3

0 a 2a

π θ − = 3

40. r a cos , r a sin 41. (a) r a cos2 tan 1, 4 r3 ar2 cos2 1 4 x2 y232 ax2 A 2 a sin 2 d 2 0 (b) 4 41 cos 2 a =4 a =6 a2 d 0 2 −66 1 sin 2 4 a2 2 2 0 −4

2 1 2 1 a 1 2 2 2 2 2 4 2 (c) A 4 6 cos 4 cos d 2 0 1 1 2 a2 a2 8 4 40 cos4 d 0 π 2 2 r = a sin θ a 10 1 cos 22 d 0 2 1 cos 4 10 1 2 cos 2 d 0 0 2 a 3 1 2 15 10 sin 2 sin 4 r = a cos θ 2 8 0 2 430 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

42. By symmetry,A1 A2 and A3 A4. 1 6 1 4 2 2 2 2 A1 A2 2a cos a d 2a cos 2a sin d 2 3 2 6 a2 6 4 4 cos2 1 d 2a2 cos 2 d 2 3 6 a2 6 4 a2 3 sin 2 a2sin 2 3 a21 a2 1 2 3 6 2 2 2 4 1 a2 A A a2 π π 3 4 2 θ = 2 2 4 3 π θ 2a = 4 1 5 1 ra= 2 sin θ 2 2 A π A a 2 2a sin d 2 5 5π A7 θ 2 6 2 θ a = 5 6 = 6 6 2 A 5a 3 A A 2 6 1 2a 1 cos 2 d 0 12 56 A a 2a A5 4 5a2 5a2 3 3 r = 2 ra= 2 cos θ a22 sin 2 a2 a2 12 56 12 3 2 12 2 π θ − = 3 1 6 1 4 2 2 A6 2 2a sin d 2 a d 2 0 2 6 6 4 2a2 1 cos 2 d a2 0 6 6 a2 3 a2 5 3 a22 sin 2 a2 a2 0 12 3 2 12 12 2 1 4 2 2 A7 2 2a sin a d 2 6 4 4 3 a2 4 sin2 1 d a2 sin 2 a2 1 6 6 12 2 2 [Note: A1 A6 A7 A4 a area of circle of radius a]

43. r a cosn For n 1: For n 2: r a cos r a cos 2 a 2 a2 1 4 a2 A A 8 a cos 22 d 2 4 2 0 2

π π 2 2

0 0 a a

—CONTINUED— Section 10.5 Area and Arc Length in Polar Coordinates 431

43. —CONTINUED— For n 3: For n 4: r a cos 3 r a cos 4 1 6 a2 1 8 a2 A 6 a cos 32 d A 16 a cos 42 d 2 0 4 2 0 2

π π 2 2

0 a 0 a

In general, the area of the region enclosed by r a cosn for n 1, 2, 3, . . . is a24 if n is odd and is a22 if n is even.

y 44. r sec 2 cos , < < 2 2 1 r cos 1 2 cos2

r2 cos2 x2 x x 1 2 1 2 r2 x2 y 2

x2 y 2x x2 y 2 2x2 −1 y 2x 1 x2 x3 x21 x y 2 1 x 1 4 A 2 sec 2 cos 2 d 2 0 4 4 4 sec2 4 4 cos2 d sec2 4 21 cos 2 d tan 2 sin 2 2 0 0 0 2

45. r a 46. r 2a cos r 0 r 2a sin

2 2 2 s a2 02 d a 2a s 2a cos 2 2a sin 2 d 0 0 2 2 (circumference of circle of radius a ) 2 2a d 2a 2a 2 2 47. r 1 sin r cos

32 s 2 1 sin 2 cos 2 d 2 32 22 1 sin d 2 3 2 cos 22 d 2 1 sin 32 42 1 sin 2 42 2 0 8 432 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

48. r 81 cos , 0 ≤ ≤ 2 r 8 sin s 2 81 cos 2 8 sin 2 d 0 16 1 2 cos cos2 sin2 d 0 162 1 cos d 0 1 cos 162 1 cos d 0 1 cos sin 162 d 0 1 cos 3221 cos 0 64

1 49. r 2, 0 ≤ ≤ 50. r sec , 0 ≤ ≤ 51. r , ≤ ≤ 2 2 3

4 3 0.5

−3 4 −0.5 0.5

−12

−1 −3 −0.5

Length 4.16 Length 1.73 exact 3 Length 0.71

52. r e, 0 ≤ ≤ 53. r sin3 cos , 0 ≤ ≤ 54. r 2 sin2 cos , 0 ≤ ≤

10 1 2

−1 2 −2 4 −25 5

−5 −1 −2

Length 31.31 Length 4.39 Length 7.78

55. r 6 cos 56. r a cos r 6 sin r a sin

2 2 S 2 6 cos sin 36 cos2 36 sin2 d S 2 a cos cos a2 cos a2 sin2 d 0 0 2 2 2 72 sin cos d 2 a2 cos2 d a2 1 cos 2 d 0 0 0 2 2 2 2 sin 2 2 a 36 sin a2 0 2 0 2 36 Section 10.5 Area and Arc Length in Polar Coordinates 433

57. r ea r aea

2 S 2 ea cos ea2 aea2 d 0 2 21 a2 e2a cos d 0 e2a 2 2 2 1 a 2 2a cos sin 4a 1 0 21 a2 ea 2a 4a2 1

58. r a1 cos r a sin S 2 a1 cos sin a21 cos 2 a2 sin2 d 2a2 sin 1 cos 2 2 cos d 0 0 42a2 32a2 22a2 1 cos 32sin d 1 cos 52 0 5 0 5

59. r 4 cos 2 60. r r 8 sin 2 r 1

4 S 2 4 cos 2 sin 16 cos2 2 64 sin2 2 d S 2 sin 2 1 d 42.32 0 4 32 cos 2 sin cos2 2 4 sin2 2 d 21.87 0

1 1 61. Area f 2 d r2 d 62. The curves might intersect for different values of : 2 2 See page 741. dr 2 Arc length f 2 f2 d r2 d d

63. (a) is correct: s 33.124. 64. (a) S 2 f sin f 2 f2 d (b) S 2 f cos f 2 f2 d

65. Revolve r 2 about the line r 5 sec . π 2 f 2, f 0 r = 5sec

2 S 2 5 2 cos 22 02 d r = 2 5 − 2cos 0

2 0 5 4 5 2 cos d 0 2 45 2 sin 0 402 434 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

π 66. Revolve r a about the line r b sec where b > a > 0. 2

f a 2a

f 0 a

2 2 2 0 S 2 b a cos a 0 d a b 0 2 2ab a sin 0 2a2b 4 2ab

67. r 8 cos , 0 ≤ ≤ 1 1 1 cos 2 sin 2 (a) A r2 d 64 cos2 d 32 d 16 16 2 0 2 0 0 2 2 0 (Area circle r2 42 16 ) (b) (c), (d) For 1 of area 4 12.57: 0.42 0.2 0.4 0.6 0.8 1.0 1.2 1.4 4 For 1 of area 8 25.13: 1.57 2 A 6.32 12.14 17.06 20.80 23.27 24.60 25.08 2 3 For 4 of area 12 37.70 : 2.73 (e) No, it does not depend on the radius.

68. r 3 sin , 0 ≤ ≤ 1 9 9 9 1 9 (a) A r2 d sin2 d 1 cos 2 d sin 2 2 0 2 0 4 0 4 2 0 4 3 3 2 Note: radius of circle is ⇒ A 2 2 (b) 0.2 0.4 0.6 0.8 1.0 1.2 1.4 A 0.0119 0.0930 0.3015 0.6755 1.2270 1.9401 2.7731

1 1 9 (c), (d) For 8 of area 8 4 0.8836 : 0.88 1 1 9 For 4 of area 4 4 1.7671 : 1.15 1 1 9 For 2 of area 2 4 3.5343 : 2 1.57

69. r a sin b cos r2 ar sin br cos x2 y2 ay bx x2 y2 bx ay 0 represents a circle. Section 10.5 Area and Arc Length in Polar Coordinates 435

70. r sin cos , Circle Converting to rectangular form: 1 r2 r sin r cos A r2 d 2 x2 y2 y x 1 sin cos 2 d 1 1 1 2 x2 x y2 y 0 4 4 2 1 1 2 sin cos d 1 2 1 2 1 2 x y 0 2 2 2 1 sin2 1 1 1 2 0 2 Circle of radius and center , 2 2 2 1 2 Area 2 2

71. (a) r , ≥ 0 12 As a increases, the spiral opens more rapidly. If < 0, the spiral is reflected about the y-axis. −10 14 (b)r a, ≥ 0, crosses the polar axis for n, n and integer. To see this

r a ⇒ r sin y a sin 0 −12 for n. The points are r, an, n, n 1, 2, 3, . . .. 1 (c) f , f 1 (d) A r2 dr 2 2 1 2 s 2 1 d lnx2 1 x xx2 1 1 2 0 2 0 2 d 2 0 1 ln42 1 2 42 1 21.2563 3 2 4 2 3 6 0 3

π 72. r e 6 2 2 = −2π 1 2 1 0 62 62 = 0 A e d e d = 2π 2 0 2 2 0 −2 123 1 2 1 0 e3 d e3 d 2 0 2 2 −3 3 2 3 0 e3 e3 2 0 2 2 3 3 3 3 3 e23 e23 e23 e23 2 2 2 2 2 2 9.3655

y 2 73. The smaller circle has equation r a cos . 2 1 = This equals the area of the square, . r 1 π θ = 2 2 The area of the shaded lune is: 4 2 a 1 1 1 4 2 x 2 4 2 4 2 A 2 a cos 1 d 2 2 0 2 2 2 a 2a 2 4 0 4 2 a r = a cos θ 1 cos 2 1 d 4 2 0 2 a2 2 2 2 4 a sin 2 2 2 0 a 2 a2 1 Smaller circle: r 2 cos 2 4 2 4 436 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

3t 3t2 74. x , y 1 t3 1 t3 27t3 t6 27t3 1 2 3 (a)x3 y3 (b)2 (c) A r2 d 33 32 1 t 1 t 2 0 2

27t3 −3 3 3xy 1 t32 Hence, x3 y3 3xy. −2 r cos 3 r sin 3 3r cos r sin 3 cos sin r cos3 sin3

75. False. f 1 and g 1 have the same graphs. 76. False. f 0 and g sin 2 have only one point of intersection.

77. In parametric form, b dx 2 dy 2 s dt. a dt dt Using instead of t, we have x r cos f cos and y r sin f sin . Thus, dx dy f cos f sin and f sin f cos . d d It follows that dx 2 dy 2 f2 f2. d d Therefore, s f 2 f2 d.

Section 10.6 Polar Equations of Conics and Kepler’s Laws

2e 1. r 4 1 e cos e = 1.5 e = 0.5 2 −4 8 (a)e 1, r , parabola 1 cos e = 1.0 1 2 (b)e 0.5, r , ellipse −4 1 0.5 cos 2 cos 3 6 (c)e 1.5, r , hyperbola 1 1.5 cos 2 3 cos

2e 2. r 4 1 e cos e = 1.5 e = 1 2 (a)e 1, r , parabola −9 3 1 cos e = 0.5 1 2 − (b)e 0.5, r , ellipse 4 1 0.5 cos 2 cos 3 6 (c)e 1.5, r , hyperbola 1 1.5 cos 2 3 cos Section 10.6 Polar Equations of Conics and Kepler’s Laws 437

2e 3. r 1 e sin e = 1.0 4 e = 0.5 2 (a)e 1, r , parabola − 1 sin 99 e = 1.5 1 2 (b)e 0.5, r , ellipse 1 0.5 sin 2 sin −8 3 6 (c)e 1.5, r , hyperbola 1 1.5 sin 2 3 sin

2e 4. r 1 e sin 9

2 (a)e 1, r , parabola e = 1.5 1 sin e = 1 −9 9 1 2 (b)e 0.5, r , ellipse 1 0.5 sin 2 sin e = 0.5 −3 3 6 (c)e 1.5, r , hyperbola 1 1.5 sin 2 3 sin

4 5. r 1 e sin

(a)5 e = 0.1 The conic is an ellipse. (c)80 e = 1.1 The conic is a hyper- − 30 30 As e → 1 , the ellipse e = 1.5 bola. As e → 1 , the e = 0.25 becomes more elliptical, e = 2 hyperbolas opens e = 0.5 → and as e 0 , it becomes − 90 90 more slowly, and as e = 0.75 more circular. e → , they open e = 0.9 − 40 − 40 more rapidly.

(b)5 The conic is a parabola. − 30 30

− 40

4 6. r 1 0.4 cos

(a) Because e 0.4 < 1, the conic is an ellipse with (c) 7 9 vertical directrix to the left of the pole. 4 −10 10 (b) r −8 8 1 0.4 cos The ellipse is shifted to the left. The vertical directrix is to −7 −5 the right of the pole 4 r . 1 0.4 sin The ellipse has a horizontal directrix below the pole.

7. Parabola; Matches (c) 8. Ellipse; Matches (f) 9. Hyperbola; Matches (a)

10. Parabola; Matches (e) 11. Ellipse; Matches (b) 12. Hyperbola; Matches (d) 438 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

1 6 13. r 14. r 1 sin 1 cos Parabola because e 1; d 1 Parabola because e 1; d 6 Distance from pole to directrix: d 1 Directrix: x 6 Directrix: y 1 Distance from pole to directrix: d 6 1 3 π Vertex: r, 3, 0 π Vertex: r, , 2 2 2 2

0 1 2 0 8 4 4 8

6 3 5 1 15. r 16. r 2 cos 1 12 cos 5 3 sin 1 35 sin

Ellipse because e 12; d 6 Ellipse because e 35 < 1; d 53 Directrix: x 6 Directrix: y 53 Distance from pole to directrix: d 6 Distance from pole to directrix: d 53 Vertices: r, 2, 0, 6, Vertices: r, 58, 2, 52, 32

π π 2 2

0 −2 2

0 1 3

17. r2 sin 4 18. r3 2 cos 6 4 2 6 2 r r 2 sin 1 12 sin 3 2 cos 1 23 cos

Ellipse because e 12; d 4 Ellipse because e 23 < 1; d 3 Directrix: y 4 Directrix: x 3 Distance from pole to directrix: d 4 Distance from pole to drectrix: d 3 Vertices: r, 43, 2, 4, 32 Vertices: r, 6, 0, 65,

π π 2 2

0 143 0 −242 Section 10.6 Polar Equations of Conics and Kepler’s Laws 439

5 5 6 2 19. r 20. r 1 2 cos 1 2 cos 3 7 sin 1 73 sin

Hyperbola because e 2 > 1; d 52 Hyperbola because e 73 > 1; d 67 Directrix: x 52 Directrix: y 67 Distance from pole to directrix: d 52 Distance from pole to directrix: d 67 Vertices: r, 5, 0, 53, Vertices: r, 35, 2, 32, 32

π π 2 2

0 −2 2

0 4 68

3 32 4 21. r 22. r 2 6 sin 1 3 sin 1 2 cos

Hyperbola because e 3 > 0; d 12 Hyperbola because e 2 > 1; d 2 Directrix: y 12 Directrix: x 2 Distance from pole to directrix: d 12 Distance from pole to directrix: d 2 Vertices: r, 38, 2, 34, 32 Vertices: r, 43, 0, 4,

π π 2 2

0 0 1 1352

23. r 34 2 sin 24. r 32 4 sin 25. r 11 cos 26. r 22 3 sin

1 4 2 4

−22 − −6 6 21−6 6

− − 2 −4 2 −4

Ellipse Hyperbola Parabola Hyperbola

1 6 2 6 27. r 28. r 1 sin −2 10 1 cos 4 3 −14 10 Rotate the graph of Rotate the graph of

1 − 6 r 6 r −10 1 sin 1 cos counterclockwise through the angle . counterclockwise through the angle . 4 3 440 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

6 6 29. r 5 30. r 4 3 7 sin 23 2 cos 6 Rotate graph of −6 6 Rotate the graph of −8 4 6 r 6 3 7 sin −4 r −3 2 cos clockwise through angle of 23. clockwise through the angle . 6

5 2 31. Change to : r 32. Change to : r 4 6 5 3 cos 1 sin 4 6

33. Parabola 34. Parabola 35. Ellipse e 1, x 1, d 1 e 1, y 1, d 1 1 e , y 1, d 1 2 ed 1 ed 1 r r 1 e cos 1 cos 1 e sin 1 sin ed r 1 e sin 12 1 12 sin 1 2 sin

36. Ellipse 37. Hyperbola 38. Hyperbola 3 e 2, x 1, d 1 3 e , y 2, d 2 e , x 1, d 1 4 2 ed 2 r ed 1 e cos 1 2 cos ed r r 1 e sin 1 e cos 234 32 1 34 sin 1 32 cos 6 3 4 3 sin 2 3 cos

39. Parabola 40. Parabola 41. Ellipse Vertex: 5, Vertices: 2, 0, 8, Vertex: 1, 2 e 1, d 10 3 16 e , d 2 5 3 e 1, d 2, r ed 10 1 sin r 1 e cos 1 cos ed r 1 e cos 165 1 35 cos 16 5 3 cos Section 10.6 Polar Equations of Conics and Kepler’s Laws 441

42. Ellipse 43. Hyperbola 44. Hyperbola 3 3 3 Vertices: 2, 0, 10, 0 Vertices: 2, , 4, Vertices: 1, , 9, 2 2 2 2 3 10 e , d 1 5 9 2 3 e , d 8 e , d 3 4 5 ed r ed ed 1 e cos r r 1 e sin 1 e sin 5 83 94 1 32 cos 1 13 sin 1 54 sin 10 8 9 2 3 cos 3 sin 4 5 sin

45. Ellipse if 0 < e < 1, parabola if e 1, hyperbola if e > 1.

4 46. r is a parabola with horizontal directrix above the pole. 1 sin (a) Parabola with vertical directrix to left of pole. (b) Parabola with horizontal directrix below pole. (c) Parabola with vertical directrix to right of pole. (d) Parabola (b) rotated counterclockwise 4.

47. (a) Hyperbola e 2 > 1 48. If the foci are fixed and e → 0, then d → . To see this, compare the ellipses 1 (b) Ellipse e < 1 10 12 r , e 12, d 1 1 12 cos (c) Parabola e 1 516 (d) Rotated hyperbola e 3 r , e 14, d 54. 1 14 cos

x2 y2 x2 y2 49. 1 50. 1 a2 b2 a2 b2 x2b2 y2a2 a2b2 x2b2 y2a2 a2b2 b2r2 cos2 a2r2 sin2 a2b2 b2r2 cos2 a2r2 sin2 a2b2 r2b2 cos2 a21 cos2 a2b2 r2b2 cos2 a21 cos2 a2b2 r2a2 cos2 b2 a2 a2b2 r2a2 cos2a2 b2 a2b2 a2b2 a2b2 a2b2 b2 r2 r2 a2 b2 a2 cos2 a2 c2 cos2 a2 c2 cos2 1 c2a2 cos2 b2 b2 b2 1 ca2 cos2 1 e2 cos2 1 e2 cos2

4 5 5 51. a 5, c 4, e , b 3 52. a 4, c 5, b 3, e 53. a 3, b 4, c 5, e 5 4 3 9 9 16 r2 r2 r2 1 1625 cos2 1 2516 cos2 1 259 cos2

3 2 54. a 2, b 1, c 3, e 1 3 2 55. A 2 d 2 0 2 cos 1 2 1 r 2 9 d 10.88 1 34 cos 2 0 2 cos 442 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

1 2 2 2 2 1 56. A 2 d 4 d 3.37 2 2 2 3 2 sin 2 3 2 sin

57. Vertices: (123,000 4000, 0 127,000, 0 ed 58. (a) r 1 e cos 119 4000, 4119, When 0, r c a ea a a1 e. 127,000 4119 a 65,559.5 2 Therefore, ed c 65,559.5 4119 61,440.5 a1 e 1 e c 122,881 e 0.93717 a 131,119 a1 e1 e ed a1 e2 ed. ed r 1 e cos 1 e2a Thus, r . 1 e cos ed ed 0: r , : r 1 e 1 e (b) The perihelion distance is a c a ea a1 e. ed ed 1 e2a 2 a 2 65,559.5 When , r a1 e. 1 e 1 e 1 e e e 2e 131,119 d d The aphelion distance is a c a ea a1 e. 1 e 1 e 1 e2 1 e2a 131,1191 e2 When 0, r a1 e. d 8514.1397 1 e 2e 7979.21 1,046,226,000 r 1 0.93717 cos 131,119 122,881 cos When 60 , r 15,015. 3 Distance between earth and the satellite is r 4000 11,015 miles.

59. a 1.496 108, e 0.0167 60. a 1.427 109, e 0.0542 1 e2a 149,558,278.1 1 e2a 1,422,807,988 r r 1 e cos 1 0.0167 cos 1 e cos 1 0.0542 cos Perihelion distance: a1 e 147,101,680 km Perihelion distance: a1 e 1,349,656,600 km Aphelion distance: a1 e 152,098,320 km Aphelion distance: a1 e 1,504,343,400 km

61. a 5.906 109, e 0.2488 62. a 5.791 107, e 0.2056 1 e2a 5,540,410,095 1 e2a 55,462,065.54 r r 1 e cos 1 0.2488 cos 1 e cos 1 0.2056 cos Perihelion distance: a1 e 4,436,587,200 km Perihelion distance a1 e 46,003,704 km Aphelion distance: a1 e 7,375,412,800 km Aphelion distance a1 e 69,816,296 km Section 10.6 Polar Equations of Conics and Kepler’s Laws 443

5.540 109 63. r 1 0.2488 cos 1 9 (a) A r2 d 9.368 1018 km2 (c) For part (a) 2 0 9 9 1 s r2 drd2 d 2.563 109 km 2 r d 0 2 0 248 21.89 yrs 2 9 1 2.563 10 8 r2 d Average per year 1.17 10 kmyr 2 0 21.89 1 For part (b) (b) r2 d 9.368 108 2 0.9018 s r2 drd2 d 4.133 109 By trial and error, 0.9018

0.9018 > 9 0.3491 because the rays in 4.133 109 Average per year 1.89 108 kmyr part (a) are longer than those in part (b). 21.89

1 64. a 250 125 au, e 0.995 2 c 1 e2a 1.246875 (a) e ⇒ c 124.375 (b) r a 1 e cos 1 0.995 cos b2 a2 c2 ⇒ b 12.4844 (c) Perihelion distance: a1 e 0.625 au length of minor axis: 2b 24.97 au Aphelion distance: a1 e 249.375 au

65. r1 a c, r0 a c, r1 r0 2c, r1 r0 2a 66. For a hyperbola, r r c 1 0 r0 c a and r1 c a. e a r1 r0 Thus r1 r0 2c and r1 r0 2a. c 1 c r r 1 e a a c r e 1 0 1 a r r 1 e c a c r 1 0 1 0 r a e 1 c a 1 c a 1 e 1 c a 1 c a r0

ed ed 67. r and r 1 1 sin 2 1 sin Points of intersection: ed, 0, ed, ed ed cos cos sin dy 1 sin 1 sin2 r : 1 dx ed ed cos sin cos 1 sin 1 sin2 dy dy At ed, 0, 1. At ed, , 1. dx dx ed ed cos cos sin dy 1 sin 1 sin2 r : 2 dx ed ed cos sin cos 1 sin 1 sin2 dy dy At ed, 0, 1. At ed, , 1. dx dx Therefore, at ed, 0 we have m1m2 1 1 1, and at ed, we have m1m2 1 1 1. The curves intersect at right angles. 444 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

c d 68. r , r (Parabolas) 1 1 cos 2 1 cos To find the intersection points: c d 1 cos 1 cos c c cos d d cos c d cos c d c cc d c d r r 1 c d 2c 2 2 1 c d dr c sin dr d sin 1 , 2 d 1 cos 2 d 1 cos 2 For the first parabola, r cos r sin 2 dy 1 1 c cos 1 cos c sin 1 cos . dx r1 sin r 1 cos c sin 1 cos c sin cos sin Similarly for the second parabola, dy sin . dx 1 cos Since the product of the slopes is 1, they intersect at right angles.

Review Exercises for Chapter 10

1. 4x2 y2 4 2. 4x2 y2 4 3. y2 4x Ellipse Hyperbola Parabola opening to left. Vertex: 1, 0. Vertex: 1, 0 Matches (b) Matches (e) Matches (c)

4. y2 4x2 4 5. x2 4y2 4 6. x2 4y Hyperbola Ellipse Parabola opening upward. Vertex: 0, 2 Vertex: 0, 1 Matches (f) Matches (d) Matches (a)

7. 16x2 16y2 16x 24y 3 0 8. y2 12y 8x 20 0 1 3 9 3 1 9 y2 12y 36 8x 20 36 x2 x y2 y 4 2 16 16 4 16 y 62 42x 2 1 2 3 2 Parabola y x y 1 2 4 Vertex: 2, 6 16

y Circle 12 1 3 Center: , 1 2 4 x Radius: 1 1 1 2 x −4 812 1 3 ,, 2 2 4 Review Exercises for Chapter 10 445

9. 3 x2 2y2 24x 12y 24 0 10. 4 x2 y2 16x 15 0 3 x2 8x 16 2y2 6y 9 24 48 18 4 x2 4x 4 y2 15 16 2 2 x 42 y 32 x 2 y y 1 1 2 3 1 4 1 1 Hyperbola y Ellipse (2, 0) Center: 2, 0 x Center: 4, 3 6 1 2 3 Vertices: 2, ±1 ± Vertices: 4 2, 3 4 −1

Asymptotes: 2 −2

3 x y 3 ± x 4 6 4 2 2

11. 3 x2 2y2 12x 12y 29 0 3 x2 4x 4 2y2 6y 9 29 12 18 x 22 y 32 1 y 13 12 x Ellipse 1 1 2 3 1 Center: 2, 3 − 2 (2, 3) 2 Vertices: 2, 3 ± 3 2 4

12. 4 x2 4y2 4x 8y 11 0 13. Vertex: 0, 2 1 Directrix: x 3 4 x2 x 4y2 2y 1 11 1 4 4 Parabola opens to the right. x 122 y 12 p 3 1 2 2 y 22 43x 0 Hyperbola y y2 4y 12x 4 0 1 4 Center: , 1 3 2

1 1 Vertices: ± 2, 1 2 x −2134 −1 1 Asymptotes: y 1 ± x −2 2

14. Vertex: 4, 2 15. Vertices: 3, 0, 7, 0 Focus: 4, 0 Foci: 0, 0, 4, 0 Parabola opens downward. Horizontal major axis p 2 Center: 2, 0 x 42 42y 2 a 5, c 2, b 21 2 x 8x 8y 0 x 22 y2 1 25 21 446 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

16. Center: 0, 0 17. Vertices: ±4, 0 Solution points: 1, 2, 2, 0 Foci: ±6, 0 Substituting the values of the coordinates of the given Center: 0, 0 points into Horizontal transverse axis x2 y2 1, a 4, c 6, b 36 16 25 b2 a2 x2 y2 we obtain the system 1 16 20 1 4 4 1, 1. b2 a2 b2 Solving the system, we have 16 x2 3y2 a2 and b2 4, 1. 3 4 16

x2 y2 5 18. Foci: 0, ±8 19. 1, a 3, b 2, c 5, e 9 4 3 Asymptotes: y ±4x By Example 5 of Section 10.1, Center: 0, 0 2 5 C 12 1 sin2 d 15.87. Vertical transverse axis 0 9 c 8 a y x 4x asymptote → a 4b b b2 c2 a2 64 4b2 ⇒ 17b2 64 64 1024 ⇒ b2 ⇒ a2 17 17 y2 x2 1 102417 6417

x2 y2 21 20. 1, a 5, b 2, c 21, e 21. y x 2 has a slope of 1. The perpendicular slope is 1. 4 25 5 y x2 2x 2 By Example 5 of Section 10.1, dy 1 5 2 21 2x 2 1 when x and y . C 20 1 sin2 d 23.01. dx 2 4 0 25 5 1 Perpendicular line: y 1x 4 2 4 x 4y 7 0

1 1 2 22. 2x y 5 has slope 2. The perpendicular slope is . 23. y x 2 200 2 y 3x2 x 6 Parabola (a) x 200y 2 1 x 4 50 y y 6x 1 2 Focus: 0, 50 1 1 6 x (b) y x2 2 200 1 95 x , y 1 12 16 y x 100 95 1 1 Perpendicular line: y x x2 16 2 12 1 y2 1 10,000

1 287 100 y x x2 2 48 S 2 x1 dx 38,294.49 0 10,000 Review Exercises for Chapter 10 447

24. (a) V abLength 1216 192 ft3 3 4 8 3 3 (b) F 262.4 3 y 9 y2 dy 62.43 9 y2 dy y9 y2 dy 3 3 3 3 3 8 3 y 1 3 62.4 y9 y2 9 arcsin 9 y232 3 2 3 3 3 8 3 9 3 9 8 27 62.4 62.4 7057.274 3 2 2 2 2 3 2 3 y (c) You want 4 of the total area of 12 covered. Find h so that Area of filled 4 xy= 9 − 2 4 h 4 3 tank above 2 9 y2 dy 3 x-axis is 3π . 3 2 0 h h 2 9 x 9 y dy −2 2 0 8 −2 1 y h 9 Area of filled 2 − tank below y 9 y 9 arcsin 4 x-axis is 6π . 2 3 0 8 h 9 h9 h2 9 arcsin . 3 4 By Newton’s Method,h 1.212. Therefore, the total height of the water is 1.212 3 4.212 ft. (d) Area of ends 212 24 Area of sides PerimeterLength

2 7 16 1 sin2 d16 from Example 5 of Section 9.1 0 16 353.656 Total area 24 353.656 429.054

25. x 1 4t, y 2 3t 26. x t 4, y t2 27. x 6 cos , y 6 sin x 1 x 1 t x 4 ⇒ y x 42 x 2 y 2 t ⇒ y 2 3 1 4 4 6 6 Parabola 3 11 x2 y2 36 y x y 4 4 7 Circle 6 4 y 3x 11 0 5 y 4 Line 3 2 4 y 1 2 4 x −11234567 x −4 −2 2 4 −2 2 −4 1 x −1 1 2 3 5 −1

−2 448 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

28. x 3 3 cos , y 2 5 sin 29. x 2 sec , y 3 tan x 3 2 y 2 2 x 22 sec2 1 tan2 1 y 32 1 y 3 5 7 x 22 y 32 1 y 6 x 32 y 22 5 8 1 4 Hyperbola 9 25 3 2 4 1 2 Ellipse x − − 2 112345678 x −2 −4 2 4 8 − −3 2 −4

30. x 5 sin3 , y 5 cos3 y 31. x 3 3 2t 3 5t 6 x 23 y 23 y 2 2 6t 2 4t 1 4 5 5 2 (other answers possible) 23 23 23 x x y 5 −6 −4 246

−4

−6

x 32 y 42 32. x h2 y k2 r2 33. 1 16 9 x 52 y 32 22 4 x 32 y 42 Let cos2 and sin2 . x 5 2 cos t, y 3 2 sin t 16 9 Then x 3 4 cos and y 4 3 sin.

y2 x2 34. a 4, c 5, b2 c2 a2 9, 1 35. x cos 3 5 cos 5 16 9 y sin 3 5 sin y2 x2 Let sec2 and tan2 . −7 8 16 9 Then x 3 tan and y 4 sec . −5

36. (a) x 2 cot , y 4 sin cos , 0 < < 37. x 1 4t

4 y 2 3t dy 3 (a) −12 12 dx 4 No horizontal tangents −4 x 1 (b) t (b) 4 x2y 4 4 cot2 4 sin cos 4 2 3 3x 11 16 csc sin cos y 2 x 1 4 4 cos 16 y sin (c) 5 82 cot 4 8x

x 1 2 3 5 Review Exercises for Chapter 10 449

1 38. x t 4 39. x t y t2 y 2t 3 dy 2t (a) 2t 0 when t 0. dy 2 dx 1 (a) 2t2 dx 1t2 Point of horizontal tangency: 4, 0 No horizontal tangents, t 0 (b) t x 4 1 (b) t y x 42 x

5 y 4 (c)

3 6 2

1 4

x 132 4 5 6 2

x 442 2

1 1 40. x 41. x t 2t 1 y t2 1 y t2 2t dy 2t (a) 2t3 dx 1t2 2t 2 dy t2 2t2 t 12t 12 No horizontal tangents, t 0 (a) 0 when t 1. dx 2 t2t 22 1 2t 12 (b) t x 1 Point of horizontal tangency: , 1 1 3 y x2 1 1 1 (b) 2t 1 ⇒ t 1 x 2 x (c) y

4 1 y 1 1 x 1 1 x 3 2 2 x 2 x 2 4x2 4x2 , x 0 1 1 x2 4x1 x 5x 1x 1

x y −2 −1 1 2 (c) 3

x −2 −123 −1

−2 450 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

42. x 2t 1 43. x 3 2 cos 1 y 2 5 sin y t2 2t dy 5 cos 3 (a) 2.5 cot 0 when , . dy t2 2t22t 2 dx 2 sin 2 2 (a) dx 2 Points of horizontal tangency: 3, 7, 3, 3 1 t 0 when t 1. x 32 y 22 t2t 22 (b) 1 4 25 Point of horizontal tangency: 1, 1 (c) y x 1 8 (b) t 2 4 1 4 y 2 x 12 2x 12 x 3x 1 x 48 (c) y 4 2

x −2 2 4

44. x 6 cos 45. x cos3 y 6 sin y 4 sin3 dy 6 cos 3 dy 12 sin2 cos (a) cot 0 when , . (a) dx 6 sin 2 2 dx 3 cos2 sin Points of horizontal tangency: 0, 6, 0, 6 4 sin 4 tan 0 cos x 2 y 2 (b) 1 6 6 when 0, . (c) y dy dx But, 0 at 0, . Hence no points of 4 dt dt 2 horizontal tangency. x −4 −2 2 4 23 −2 y (b) x23 1 −4 4

x 42 2 4

4 Review Exercises for Chapter 10 451

46. x et 47. x 4 t, y t2 y et dx dy 1, 2t dt dt dy et 1 1 (a) dx et e2t x2 dy 0 for t 0. dt No horizontal tangents Horizontal tangent at t 0: x, y 4, 0 (b) t ln x No vertical tangents 1 y eln x eln1x , x > 0 x (c) y

x 1 2 3

48. x t 2, y t3 2t dx dy 1, 3t2 2 dt dt dy 2 ±6 0 for t ± dt 3 3 2 6 26 2 Horizontal tangents: t : x, y 2, 6 3 3 9 3 2.8165, 1.0887 6 6 2 26 t : x, y 2, 6 3 3 3 9 1.1835, 1.0887 No vertical tangents

49. x 2 2 sin , y 1 cos 50. x 2 2 cos , y 2 sin 2 dx dy dx dy 2 cos , sin 2 sin , 4 cos 2 d d d d dy dy 3 5 7 0 for 0, , 2, . . . 0 for , , , , . . . d d 4 4 4 4 Horizontal tangents: x, y 2, 2, 2, 0 Horizontal tangents: x, y 2 ± 2, 2, dx 3 2 ± 2, 2 0 for , , . . . d 2 2 dx 0 for 0, , 2 , . . . Vertical tangents: x, y 4, 1, 0, 1 d Vertical tangents: x, y 0, 0, 4, 0 452 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

51. x cot 52. x 2 sin y sin 2 2 sin cos y 2 cos

(a), (c) 2 (a), (c) 8

−33 −88

−2 −4 dx dy dy 1 (b) At , 4, 1, and . dx 3 6 d d dx 4 (b) At , 1.134, 2 , 6 d 2 dy dy 0.5, and 0.441. dt dx

53. x rcos sin 54. x 6 cos y rsin cos y 6 sin dx dx r cos 6 sin d d dy dy r sin 6 cos d d

s r 2 cos2 2 sin2 d s 36 sin2 36 cos2 d 6 6 0 0 0 r 1 (one-half circumference of circle) r d 2 2r 0 2 0 2

55. x t, y 3t, 0 ≤ t ≤ 2 56. x 2 cos , y 2 sin , 0 ≤ ≤ 2 dx dy dx 2 dy 2 1, 3, 1 9 10 dt dt dt dt dx dy dx 2 dy 2 2 sin , 2 cos , 2 dt dt d d 2 t2 2 (a) S 2 3t 10 dt 6 10 12 10 2 2 2 0 0 (a) S 2 2 sin 2d 8cos 8 0 2 2 0 (b) S 2 10 dt 210 t 410 2 2 0 0 (b) S 2 2 cos 2 d 8 sin 8 0 0 1 Note: The surface is a hemisphere: 422 8 2

b 0 57. x 3 sin , y 2 cos 58. A y dx sin 2 sin d a b 2 0 A y dx 2 cos 3 cos d 1 cos 2 a 2 d 2 2 1 cos 2 0 6 d sin 2 2 2 2 sin 2 2 3 2 2 3 3 2 2 Review Exercises for Chapter 10 453

11 59. r, 3, π 60. r, 4, 2 2 6 π (3, 2 ) 11 11 x, y 3 cos , 3 sin x, y 4 cos , 4 sin 2 2 6 6 π 0, 3 2 3, 2 2 5 0 123 4 π −4, 11 3 ()6 2 1

0 −5 −4 −3 −2 −1 1 −1

61. r, 3, 1.56 62. r, 2, 2.45 x, y 3 cos1.56, 3 sin1.56 x, y 2 cos2.45, sin2.45 π π 0.0187, 1.7319 2 1.5405, 1.2755 2

3 ( 3 , 1.56) 2 (−2, −2.45) 1

0 −1123 0 12 −1

63. x, y 4, 4 64. x, y 1, 3 r 42 42 42 r 12 32 10 7 arctan3 1.89108.43 4 r, 10, 1.89, 10, 5.03 7 3 r, 42, , 42, y 4 4 (−1, 3)

y 2

1 1

x x 1 253 4 −3 −2 −1 123 −1 −1 −2 −2 −3 −3 −4 (4,− 4) −5

65. r 3 cos 66. r 10 r2 3r cos r2 100 x2 y2 3x x2 y2 100 x2 y2 3x 0

1 67. r 21 cos 68. r 2 cos r2 2r1 cos 2 r r cos 1 x2 y2 2±x2 y2 2x 2 ±x2 y2 x 1 x2 y2 2x2 4x2 y2 4 x2 y2 x 12 3 x2 4y2 2x 1 0 454 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

4 69. r2 cos 2 cos2 sin2 70. r 4 sec 3 cos 3 r 4 r2 cos2 r2 sin2 4 x2 y22 x2 y2 12 cos 32 sin rcos 3 sin 8 x 3 y 8

3 71. r 4 cos 2 sec 72. 73. x2 y22 ax2y 4 1 r4 ar2 cos2 r sin 42 cos2 1 tan 1 cos r a cos2 sin y r cos 8 cos2 4 1 x x2 x 8 4 y x x2 y2 x3 xy2 4x2 4y2 4 x y2 x2 4 x

y 2 y 2 74. x2 y2 4x 0 75. x2 y2 a2arctan 76. x2 y2arctan a2 x x r2 4r cos 0 r2 a22 r22 a2 r 4 cos

1 77. r 4 78. 79. r sec 12 cos Circle of radius 4 Line r cos 1, x 1

Centered at the pole π 2 Vertical line

Symmetric to polar axis, π 2 2, and pole

π 0 2 12

0 1

0 2 6

π 80. r 3 csc , r sin 3, y 3 81. r 2 1 cos 2 Horizontal line Cardioid

π 2 Symmetric to polar axis 0 1

0 2 1234 0 3 2 3 r 4 3 2 1 0 Review Exercises for Chapter 10 455

π π 82. r 3 4 cos 2 83. r 4 3 cos 2 Limaçon Limaçon Symmetric to polar axis 2 Symmetric to polar axis 0 0 24

2 2 0 0 3 2 3 3 2 3 5 11 r 1 13 5 7 r 14 7 2 2

π 84. r 2 2 85. r 3 cos 2 Spiral Rose curve with four petals 0 Symmetric to 2 24 8 Symmetric to polar axis, , and pole 2 3 Relative extrema: 3, 0, 3, , 3, , 3, 2 2 3 π 3 5 3 Tangents at the pole: , 2 0 4 4 4 2 4 4 2 3 5 r 03 2 2 2 2 0 4

π 86. r cos 5 2 Rose curve with five petals Symmetric to polar axis 0 2 3 4 1 Relative extrema: 1, 0, 1, , 1, , 1, , 1, 5 5 5 5 3 7 9 Tangents at the pole: , , , , 10 10 2 10 10

2 2 π 87. r 4 sin 2 2 r ±2 sin2 Rose curve with four petals 0 2 Symmetric to the polar axis, , and pole 2 3 Relative extrema: ±2, , ±2, 4 4 Tangents at the pole: 0, 2 456 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

88. r2 cos 2 π 0 2 Lemniscate 6 4 r ±1 ± 2 0 Symmetric to the polar axis 2 0 1 Relative extrema: ±1, 0 2 3 Tangents at the pole: , 4 4

3 89. r 5 90. r 2 sin cos2 0.75 cos 4 Bifolium Graph of r 3 sec rotated through an angle of 4 Symmetric to 2 −1 1 −1 8

−1 −0.25

91. r 4 cos 2 sec 4 92. r 4sec cos 3 Strophoid Semicubical parabola −66 −1 5 Symmetric to the polar axis Symmetric to the polar axis r ⇒ as ⇒ −4 r ⇒ as ⇒ −3 2 2 r ⇒ as ⇒ r ⇒ as ⇒ 2 2

93. r 1 2 cos (a) The graph has polar symmetry and the tangents at the pole are , . 3 3 dy 2 sin2 1 2 cos cos (b) dx 2 sin cos 1 2 cos sin 1 ± 1 32 1 ± 33 Horizontal tangents: 4 cos2 cos 2 0, cos 8 8 1 ± 33 1 33 3 33 When cos , r 1 2 , 8 8 4 3 33 1 33 , arccos 0.686, 0.568 4 8 3 33 1 33 , arccos 0.686, 0.568 4 8 3 33 1 33 , arccos 2.186, 2.206 4 8 3 33 1 33 , arccos 2.186, 2.206. 4 8 1 1 Vertical tangents: sin4 cos 1 0, sin 0, cos , 0, , ±arccos , 1, 0, 3, 4 4 1 1 , ±arccos 0.5, ±1.318 2 4

−5 1

−2.5 Review Exercises for Chapter 10 457

94. r2 4 sin2 dy r cos 4 cos 2 sinr (b) dx r sin 4 cos 2 cosr dr (a) 2 r 8 cos2 d cos2 sin sin2 cos cos2 cos sin2 sin dr 4 cos2 d r Horizontal tangents: dy Tangents at the pole: 0, 0 when cos 2 sin sin 2 cos 0, 2 dx (c) 2 tan tan2, 0, , 0, 0, ±23, 3 3

−3 3 Vertical tangents when cos 2 cos sin 2 sin 0: tan 2 tan 1, 0, , 0, 0, ±23, −2 6 6

95. Circle: r 3 sin dy 3 cos sin 3 sin cos sin 2 dy tan 2 at , 3 dx 3 cos cos 3 sin sin cos2 sin2 6 dx Limaçon: r 4 5 sin dy 5 cos sin 4 5 sin cos dy 3 at , dx 5 cos cos 4 5 sin sin 6 dx 9 Let be the angle between the curves: 3 39 23 tan . 1 13 3 23 Therefore, arctan 49.1. 3

96. False. There are an infinite number of polar coordinate representations of a point. For example, the point x, y 1, 0 has polar representations r, 1, 0, 1, 2, 1, , etc.

97. r 1 cos, r 1 cos The points 1, 2 and 1, 32 are the two points of intersection (other than the pole). The slope of the graph of r 1 cos is dy r sin r cos sin2 cos1 cos m . 1 dx r cos r sin sin cos sin1 cos At 1, 2 , m1 1 1 1 and at 1, 3 2 , m1 1 1 1. The slope of the graph of r 1 cos is dy sin2 cos1 cos m . 2 dx sin cos sin1 cos At 1, 2 , m2 1 1 1 and at 1, 3 2 , m2 1 1 1. In both cases,m1 1 m2 and we conclude that the graphs are orthogonal at 1, 2 and 1, 32.

98. r a sin, r a cos The points of intersection are a2, 4 and 0, 0. For r a sin, dy a cos sin a sin cos 2 sin cos m . 1 dx a cos2 a sin2 cos 2 At a 2, 4 , m1 is undefined and at 0, 0 , m1 0 . For r a cos , dy a sin2 a cos2 cos 2 m . 2 dx a sin cos a cos sin 2 sin cos At a 2, 4 , m2 0 and at 0, 0 , m2 is undefined. Therefore, the graphs are orthogonal at a 2, 4 and 0, 0 . 458 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

99. r 2 cos 100. r 51 sin 1 9 1 3 2 75 A 2 2 cos2 d 14.14, A 2 51 sin2 d 117.81, 2 0 2 2 2 2

3 4

−8 8 −36

− 3 −12

101. r2 4 sin 2 102. r 4 cos, r 2 1 2 1 3 1 2 A 2 4 sin 2 d 4 A 2 4 d 4 cos2 d 4.91 2 0 2 0 2 3

2 3

−33 −36

−2 −3

103. r sin cos2 104. r 4 sin 3 1 2 1 3 A 2 sin cos2 2 d A 3 4 sin 32 d 2 0 2 0 12.57 4 0.10, 32 4

0.5

−6 6

−0.5 0.5 −4 −0.1

105. r 3, r2 18 sin 2 4 9 r2 18 sin 2 −6 6 1 sin 2 2 −4 12 1 12 1 5 12 1 2 A 2 18 sin 2 d 9 d 18 sin 2 d 2 0 2 12 2 512 1.2058 9.4248 1.2058 11.84

106. r e, 0 ≤ ≤ 1 A e 2 d 133.62 2 0

−25 5

−5 Review Exercises for Chapter 10 459

107. r a1 cos , 0 ≤ ≤ 108. r a cos 2, 4 ≤ ≤ 2 dr dr a sin 2a sin 2 d d

2 s a21 cos 2 a2 sin2 d s a2 cos2 2 4a2 sin2 2 d 0 2 2 2a 1 cos d a 1 3 sin2 2 d 0 2 sin Using a graphing utility, s 4.8442a. 2a d 0 1 cos 22a1 cos 12 4a 0

109. f 1 4 cos f 4 sin

f2 f2 1 4 cos 2 4 sin 2 17 8 cos 2 S 2 1 4 cos sin 17 8 cos d 0 3417 88.08 5

110. f 2 sin f 2 cos f2 f2 4 sin2 4 cos2 2 2 S 2 2 sin cos 2 d 0 4

2 2 111. r , e 1 π 112. r , e 1 π 1 sin 2 1 cos 2 Parabola Parabola

0 2

0 264 8

6 2 2 4 45 3 113. r , e 114. r , e 3 2 cos 1 23 cos 3 5 3 sin 1 35 sin 5 Ellipse Ellipse

π π 2 2

0 2

0 12 460 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

4 2 3 8 4 5 115. r , e 116. r , e 2 3 sin 1 32sin 2 2 5 cos 1 52cos 2 Hyperbola Hyperbola

π π 2 2

0 234

0 12

117. Circle 118. Line Slope: 3 Center:5, 0, 5 in rectangular coordinates 2 Solution point: 0, 0 Solution point: 0, 0 y 3 x, r sin 3 r cos, 2 5 x y 5 25 tan 3, x2 y2 10y 0 3 r2 10r sin 0 r 10 sin

119. Parabola 120. Parabola Vertex: 2, Vertex: 2, 2 Focus: 0, 0 Focus: 0, 0 e 1, d 4 e 1, d 4 4 r 1 cos 4 r 1 sin

121. Ellipse 122. Hyperbola Vertices: 5, 0, 1, Vertices: 1, 0, 7, 0 Focus: 0, 0 Focus: 0, 0 2 5 4 7 a 3, c 2, e , d a 3, c 4, e , d 3 2 3 4 25 47 3 2 5 3 4 7 r r 2 3 2 cos 4 3 4 cos 1 cos 1 cos 3 3 Problem Solving for Chapter 10 461

Problem Solving for Chapter 10

1. (a) y 10

4 (4, 4) −1, 1 ( 4) 2 x −6 −4 −2 2 4 6 −2

(b) x2 4y 2 x 4y 1 y x 2 y 4 2x 4 ⇒ y 2x 4 Tangent line at 4, 4 1 1 1 1 1 y x 1 ⇒ y x Tangent line at 1, 4 2 2 4 4 Tangent lines have slopes of 2 and 12 ⇒ perpendicular. (c) Intersection: 1 1 2 x 4 x 2 4 8 x 16 2x 1 10x 15 3 3 x ⇒ , 1 2 2 Point of intersection,32, 1, is on directrix y 1.

2. Assume p > 0. y 2 Let y mx p be the equation of the focal chord. x = 4py (0, p) First find x-coordinates of focal chord endpoints: x x2 4py 4pmx p x2 4pmx 4p2 0 y = −p 4pm ± 16p2m2 16p2 x 2pm ± 2pm2 1 2 x x2 4py, 2x 4py ⇒ y . 2p (a) The slopes of the tangent lines at the endpoints are perpendicular because 1 1 1 1 2pm 2pm2 1 2pm 2pm2 1 4p2m2 4p2m2 1 4p2 1 2p 2p 4p2 4p2 —CONTINUED— 462 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

2. —CONTINUED— (b) Finally, we show that the tangent lines intersect at a point on the directrix y p. Let b 2pm 2pm2 1 and c 2pm 2pm2 1. b2 8p2m2 4p2 8p2mm2 1 c2 8p2m2 4p2 8p2mm2 1 b2 2pm2 p 2pmm2 1 4p c2 2pm2 p 2pmm2 1 4p b2 b bx b2 Tangent line at x b: y x b ⇒ y 4p 2p 2p 4p c2 c cx c2 Tangent line at x c: y x c ⇒ y 4p 2p 2p 4p bx b2 cx c2 Intersection of tangent lines: 2p 4p 2p 4p 2 bx b2 2cx c2 2 xb c b2 c2 2 x4pm2 1 16p2mm2 1 x 2pm Finally, the corresponding y -value is y p, which shows that the intersection point lies on the directrix.

3. Consider x2 4py with focus F 0, p. y B A Let Pa, b be point on parabola. F x P(a, b) 2x 4py ⇒ y 2p x Q a y b x a Tangent line at P 2p a a2 4pb For x 0, y b a b b b. 2p 2p 2p Thus, Q 0, b. FQP is isosceles because FQ p b FP a 02 b p2 a2 b2 2bp p2 4pb b2 2bp p2 b p2 b p. Thus, FQP BPA FPQ. Problem Solving for Chapter 10 463

x2 y2 4. 1, a2 b2 c2, MF MF 2a y a2 b2 2 1 β b2x M(x , y ) y 2 α 0 0 a y − β F2( c, 0) x 2 b x F1(c, 0) 0 Tangent line at M x0, y0 : y y0 2 x x0 Q a y0 yy y 2 x x x 2 0 0 0 0 b2 a2 x x y y x 2 y 2 0 0 0 0 a2 b2 a2 b2 x x y y 0 0 1 a2 b2 b2 b2 At x 0, y ⇒ Q 0, . By the Law of Cosines, y0 y0 F Q2 MF 2 MQ2 2MF MQcos b4 2 2 2 QF QF c2 d 2 1 2 2 2 2 y0 d MF2 f 2 f MF2 cos b2 2 F Q2 MF 2 f 2 2 f MF cos 2 1 1 1 MQ x0 y0 f y0 2 2 2 d MF1 f 2 f MF1 cos . F 2 f 2 d2 MF 2 f 2 d2 2 1 cos , cos 2 f MF2 2 f MF1 MF2 MF1 2a. Let z MF1. y b2 2 0 b Slopes: MF1: ; QF1: ; QF2: x0 c y0c y0c

To show , consider

2 2 2 2 2 2 MF2 f d 2 f MF1 MF1 f d 2 f MF2 ⇔ z 2a2 f 2 d 2z z2 f 2 d 2z 2a ⇔ z2 2az f 2 d2 b2 2 b4 ⇔ 2 2 2 2 x0 c y0 2az x0 y0 c 2 y0 y0 ⇔ 2 az x0c a 0 ⇔ 2 2 2 a x0 c y0 x0c a ⇔ 2 2 2 2 2 2 x0 b a y0 a b x 2 y 2 ⇔ 0 0 1. a2 b2 Thus, and the reflective property is verified. 464 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

2a 5. (a) In OCB, cos ⇒ OB 2a sec . OB OA In OAC, cos ⇒ OA 2a cos . 2a r OP AB OB OA 2asec cos 1 2a cos cos sin2 2a cos 2a tan sin (b) x r cos 2a tan sin cos 2a sin2

1 + t2 y r sin 2a tan sin sin 2a tan sin2 , < < t 2 2 θ Let t tan , < t < . 1 t2 t2 t3 Then sin2 and x 2a , y 2a . 1 t2 1 t2 1 t2 (c) r 2a tan sin r cos 2a sin2 r3 cos 2a r2 sin2 x2 y2x 2ay2 x3 y2 2a x

a b 4b 1 x a 6. (a) A 4 a2 x2 dx xa2 x2 a2 arcsin ab 0 a a 2 a 0 b a2 2a2 b 2a2 1 b 4 2 2 2 2 2 3 2 (b) Disk: V 2 2 b y dy 2 b y dy 2 b y y a b 0 b b 0 b 3 0 3 b a b4 a2 b2y2 S 4 b2 y2 dy 2 2 0 b b b y 4a b 2a b 4 2 2 4 2 2 4 4 2 2 2 b c y dy 2 cy b c y b lncy b c y b 0 b c 0 2a b2cb2 c2 b4 lncb bb2 c2 b4 lnb2 b2c ab2 c a 2 b2 1 e 2a2 ln 2a2 ln c e e 1 e a b2 2b2 a 2b2 1 a 4 2 2 2 2 2 3 2 (c) Disk: V 2 2 a x dx 2 a x dx 2 a x x ab 0 a a 0 a 3 0 3 a b a4 a2 b2x2 S 22 a2 x2 dx 2 2 0 a a a x 4b a 2b cx a 4 2 2 4 2 2 4 2 a c x dx 2 cx a c x a arcsin 2 a 0 a c a 0 ab c ab a2ca2 c2 a4 arcsin 2b2 2 arcsine a2c a e Problem Solving for Chapter 10 465

t21 t22 1 t22 1 r cos 7. (a) y2 , x2 (b) r2 sin2 r2 cos2 1 t22 1 t22 1 r cos 1 t2 sin2 1 r cos cos2 1 r cos 1 1 x 1 t2 2t2 t2 r cos sin2 sin2 cos2 r cos3 1 x 1 t2 2 1 1 t2 r cos sin2 cos2 cos2 sin2 1 x r cos cos 2 Thus, y2 x2 . 1 x r cos 2 sec

3 (c) π (d)r 0 for , . 2 4 4 Thus,y x and y x are tangent lines to curve at the origin. 0 12

1 t21 3t2 t t32t 1 4t2 t4 (e) yt 0 1 t22 1 t22 1 2 ± 5 t4 4t2 1 0 ⇒ t2 2 ± 5 ⇒ x 3 5 1 2 ± 5 1 ± 5 3 5 5 1 1 5 2 5 1 5 1 , ± 2 5 2 2

a y 8. y a1 cos ⇒ cos 9. (a) a a y arccos a x a sin a y a y aarccos sinarccos a a Generated by Mathematica a y 2ay y2 t t aarccos u2 u2 a a (b)x, y cos du, sin du is on 0 2 0 2 a y x a arccos 2ay y2, 0 ≤ y ≤ 2a a the curve whenever x, y is on the curve. t2 t2 (c) xt cos , yt sin , xt2 yt2 1 2 2 a − 2 a 2ay y Thus, s dt a. θ 0 − a y On , , s 2. 466 Chapter 10 Conics, Parametric Equations, and Polar Coordinates

3 5 7 ab 10. For t , , , , . . . 11. r , 0 ≤ ≤ 2 2 2 2 a sin b cos 2 2 2 2 2 ra sin b cos ab y , , , , . . . 3 5 7 ay bx ab Hence, the curve has length greater that y x 1 2 2 2 2 b a S . . . 3 5 7 Line segment 2 1 1 1 1 . . . 1 3 5 7 Area ab 2 2 1 1 1 1 > . . . 2 4 6 8 . (Harmonic series)

1 12. (a) A rea r2 d 13. Let r, be on the graph. 0 2 x = 1 2 2 θ r 1 2r cos r 1 2r cos 1 1 r = sec 2 sec d α r2 12 4r2 cos2 1 2 0 1 h 1 r 4 2r2 1 4r2 cos2 1 (b) tan ⇒ Area 1tan 1 2 r2r2 4 cos2 2 0 ⇒ tan sec2 d r2 4 cos2 2 0 r2 22 cos2 1 d 2 (c) Differentiating, tan sec . 2 d r 2 cos 2

14. If a dog is located at r, in the first quadrant, then its neighbor is at r, : 2 x1, y1 r cos , r sin and x2, y2 r sin , r cos . The slope joining these points is r cos r sin sin cos slope of tangent line at r, . r sin r cos sin cos dy dr sin r cos dy dr d sin cos dx dx dr sin cos cos r sin dr d dr ⇒ r d dr d r ln r C1

C r e 1 r Ce d d d r ⇒ r Ce4 ⇒ C e4 4 2 2 2 d Finally, r e4, ≥ . 2 4 Problem Solving for Chapter 10 467

15. (a) The first plane makes an angle of 70 with the positive (c) 280 x-axis, and is 150 miles from P: x1 cos 70 150 375t y1 sin 70 150 375t 0 1 0 Similarly for the second plane, The minimum distance is 7.59 miles when x2 cos 135 190 450t t 0.4145. cos 45190 450t y2 sin 135 190 450t sin 45190 450t.

2 2 (b) d x2 x1 y2 y1 cos 45190 450t cos 70150 375t2 sin 45190 450t sin 70150 375t212

16. The curve is produced over the interval 0 ≤ ≤ 10.

17. 4 4 4 4 n = −5 n = −4 − n = 3 n = −2

−6 6 −6 6 −6 6 −6 6

−4 −4 −4 −4

4 4 4 4

n = 1 n = 2 n = −1 n = 0 −6 6 −6 6 −6 6 −6 6

−4 −4 −4 −4

4 4

4 n = 4 n = 5

n = 3 −6 6 −6 6

−6 6

−4 −4

−4

n 1, 2, 3, 4, 5 produce “bells”; n 1, 2, 3, 4, 5 produce “hearts”.